Atoms JEE Advanced PYQ | Chapter-wise Physics Previous Year Questions

Q 1 :

A metal target with atomic number Z = 46 is bombarded with a high energy electron beam. The emission of X-rays from the target is analyzed. The ratio $r$ of the wavelengths of the $K_{α}$ -line and the cut-off is found to be $r = 2$ . If the same electron beam bombards another metal target with Z = 41, the value of $r$ will be [2024]

2.53
1.27
2.24
1.58

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(1)

The ratio of the wavelengths of the $K_{α}$ -line and the cut-off i.e.

$\frac{λ_{K_{α}}}{λ_{0}} (= r) \propto \frac{1}{{(Z - 1)}^{2}}$

$∴$ $\frac{r_{2}}{r_{1}} = \frac{{(Z_{1} - 1)}^{2}}{{(Z_{2} - 1)}^{2}} = \frac{{(46 - 1)}^{2}}{{(41 - 1)}^{2}} = \frac{45^{2}}{40^{2}}$

or, $r_{2} = 1.265 \times r_{1} = 1.265 \times 2 = 2.53$

Q 2 :

In a hydrogen-like atom electron make transition from an energy level with quantum number $n$ to another with quantum number $(n - 1)$ . If $n ≫ 1$ , the frequency of radiation emitted is proportional to: [2012]

$\frac{1}{n}$
$\frac{1}{n^{2}}$
$\frac{1}{n^{3} / 2}$
$\frac{1}{n^{3}}$

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(4)

$Δ E = h ν$

$ν = \frac{Δ E}{h} = k [\frac{1}{{(n - 1)}^{2}} - \frac{1}{n^{2}}] = \frac{k (2 n - 1)}{n^{2} {(n - 1)}^{2}} \approx \frac{2 k}{n^{3}}$ or, $ν \propto \frac{1}{n^{3}}$

Q 3 :

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 $Å$ . The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is [2011]

1215 $Å$
1640 $Å$
2430 $Å$
4687 $Å$

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(1)

We know for hydrogen or hydrogen-like atom,

$\frac{1}{λ} = R Z^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

For the first spectral line in the Balmer series of hydrogen atom, $n_{1} = 2$ and $n_{2} = 3$ . Here $Z = 1$

$∴$ $\frac{1}{6561} = R {(1)}^{2} (\frac{1}{4} - \frac{1}{9}) = \frac{5 R}{36}$ $\dots (i)$

For the second spectral line in the Balmer series of singly ionised helium ion, $n_{2} = 4$ and $n_{1} = 2$ $;$ $Z = 2$

$∴$ $\frac{1}{λ} = R {(2)}^{2} [\frac{1}{4} - \frac{1}{16}] = \frac{3 R}{4}$ $\dots (i i)$

Dividing Eq. (i) by Eq. (ii),

$\frac{λ}{6561} = \frac{5 R}{36} \times \frac{4}{3 R} = \frac{5}{27}$

$∴$ $λ = 1215 Å$

Q 4 :

The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is [2007]

802 nm
823 nm
1882 nm
1648 nm

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(2)

The smallest frequency and longest wavelength in the ultraviolet region will be for transition of electron from $n = 2$ to $n = 1$ , i.e., Lyman series.

$∴$ $\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

$\Rightarrow \frac{1}{122 \times 10^{- 9} m} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = R (1 - \frac{1}{4}) = \frac{3 R}{4}$

$∴$ $R = \frac{4}{3 \times 122 \times 10^{- 9}} m^{- 1}$

The highest frequency and smallest wavelength for the infrared region will be for transition of electron $n = \infty, n = 3$ corresponding to Paschen series.

$∴$ $\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$ $\Rightarrow \frac{1}{λ} = \frac{4}{3 \times 122 \times 10^{- 9}} (\frac{1}{3^{2}} - \frac{1}{\infty})$

$∴$ $λ = \frac{3 \times 122 \times 9 \times 10^{- 9}}{4} = 823.5 nm$

Q 5 :

A photon collides with a stationary hydrogen atom in the ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of microsecond another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector? [2005]

One photon of energy 10.2 eV and an electron of energy 1.4 eV
2 photon of energy of 1.4 eV
2 photon of energy 10.2 eV
One photon of energy 10.2 eV and another photon of energy 1.4 eV

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(1)

Initially, a photon of energy 10.2 eV collides inelastically with a hydrogen atom in the ground state. For hydrogen atom,

$E_{1} = - 13.6 eV, E_{2} = - \frac{13.6}{4} eV = - 3.4 eV$

$∴$ $E_{2} - E_{1} = 10.2 eV$

The electron of the hydrogen atom will jump to the second orbit after absorbing the photon of energy 10.2 eV. Another photon of energy 15 eV strikes the hydrogen atom inelastically. This energy is sufficient to knock out the electron from the atom as the ionisation energy is 13.6 eV. The remaining energy, $15 - 13.6 = 1.4 eV$ is left, which is released by the second photon.

Q 6 :

If the atom ${}_{100}{Fm}^{257}$ follows the Bohr model and the radius of ${}_{100}{Fm}^{257}$ is $n$ times the Bohr radius, then find $n$ . [2003]

$100$
$200$
$4$
$\frac{1}{4}$

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(4)

For an atom following Bohr's model, the radius of the fifth orbit.

$r_{m} = \frac{r_{0} m^{2}}{Z},$ where $r_{0}$ = Bohr's radius.

For ${}_{100}{Fm}^{257}$ , $m = 5$ (Fifth orbit in which the outermost electron is present) and $Z = 100$

$∴$ $r_{m} = \frac{r_{0} \times 5^{2}}{100} = n r_{0}$ $(given)$

$∴$ $n = \frac{1}{4}$

Q 7 :

The electric potential between a proton and an electron is given by $V = V_{0} \ln (\frac{r}{r_{0}}),$ where $r_{0}$ is a constant. Assuming Bohr's model to be applicable, write variation of $r_{n}$ with $n$ , $n$ being the principal quantum number? [2003]

$r_{n} \propto n$
$r_{n} \propto \frac{1}{n}$
$r_{n} \propto n^{2}$
$r_{n} \propto \frac{1}{n^{2}}$

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(1)

Given potential energy $(U)$ between electron and proton

$U = e V_{0} \ln (\frac{r}{r_{0}})$ $[∵ | U | = e V]$

$∴$ $| F |$ $=$ $| - \frac{d U}{d r} |$ $= \frac{d}{d r}$ $[e V_{0} \ln (\frac{r}{r_{0}})] = \frac{e V_{0}}{r_{0}} \times \frac{1}{r}$

This force will provide the necessary centripetal force.

$∴$ $\frac{m v^{2}}{r} = \frac{e V_{0}}{r r_{0}} \Rightarrow m v^{2} = \frac{e V_{0}}{r_{0}}$ $\dots (i)$

As per Bohr's postulate, $m v r = \frac{n h}{2 π}$ $\dots (i i)$

From Eqs. (i) and (ii),

$\frac{m^{2} v^{2} r^{2}}{m v^{2}} = \frac{n^{2} h^{2} r_{0}}{4 π^{2} V_{0} e}$ $\Rightarrow r^{2} = \frac{n^{2} h^{2} r_{0}}{4 π^{2} V_{0} m e}$

$∴$ $r \propto n$

Q 8 :

A Hydrogen atom and a ${Li}^{+ +}$ ion are both in the second excited state. If $ℓ_{H}$ and $ℓ_{L i}$ are their respective electronic angular momenta, and $E_{H}$ and $E_{L i}$ their respective energies, then [2002]

$ℓ_{H} > ℓ_{L i}$ and $| E_{H} | > | E_{L i} |$
$ℓ_{H} = ℓ_{L i}$ and $| E_{H} | < | E_{L i} |$
$ℓ_{H} = ℓ_{L i}$ and $| E_{H} | > | E_{L i} |$
$ℓ_{H} < ℓ_{L i}$ and $| E_{H} | < | E_{L i} |$

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(2)

$l = \frac{n h}{2 π},$ $| E | \propto \frac{Z^{2}}{n^{2}};$ $n = 3$

$\Rightarrow l_{H} = l_{Li}$ and $| E_{H} | < | E_{Li} |$

Q 9 :

The transition from the state $n = 4$ to $n = 3$ in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition. [2001]

2 → 1
3 → 2
4 → 2
5 → 4

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(4)

For state transition, 2 to 1, 3 to 2 and 4 to 2 we get energy that $n = 4$ to $n = 3$ ,

Infrared radiation has less energy and greater than ultraviolet radiation.

Infrared radiation will be obtained in the transition 5 to 4.

Q 10 :

The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true? [2000]

Its kinetic energy increases and its potential and total energies decrease.
Its kinetic energy decreases, potential energy increases and its total energy remains the same.
Its kinetic and total energies decrease and its potential energy increases.
Its kinetic, potential and total energies decrease.

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(1)

According to the question, in a hydrogen atom, the electron makes a transition from an excited state to the ground state, i.e., the electron comes closer to the nucleus, so $r$ decreases.

$∴$ $Potential Energy (P.E.) = - \frac{K Z e^{2}}{r}$

$∴$ Potential energy decreases.

Kinetic energy (K.E.) will increase.

$∵$ $K.E. = \frac{1}{2} \frac{K Z e^{2}}{r}$

$∴$ Total energy decreases.

$∵$ $T.E. = P.E. + K.E. = - \frac{1}{2} \frac{K Z e^{2}}{r}$

Q 11 :

Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength $λ$ (given in terms of the Rydberg constant $R$ for the hydrogen atom) equal to [2000]

9/(5R)
36/(5R)
18/(5R)
4/R

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(3)

For an ordinary hydrogen atom, the longest wavelength,

$\frac{1}{λ} = R [\frac{1}{2^{2}} - \frac{1}{3^{2}}] = \frac{5 R}{36}$ or $λ = \frac{36}{5 R}$

With the hypothetical particle, the required wavelength is

$λ' = \frac{1}{2} \times \frac{36}{5 R} = \frac{18}{5 R}$ $∵$ $λ \propto \frac{1}{m}$

Q 12 :

A Hydrogen-like atom has atomic number Z. Photons emitted in the electronic transitions from level $n = 4$ to level $n = 3$ in these atoms are used to perform photoelectric effect experiment on a target metal. The maximum kinetic energy of the photoelectrons generated is 1.95 eV. If the photoelectric threshold wavelength for the target metal is 310 nm, the value of Z is ______. [2023]

[Given: hc = 1240 eV–nm and Rhc = 13.6 eV, where R is the Rydberg constant, h is Planck's constant and c is the speed of light in vacuum.]

(3)

$K_{\max} = E - W$ $\Rightarrow E_{4 \to 3} = K_{\max} + W = 1.95 + \frac{h c}{λ}$

$= 1.95 + \frac{1240}{310} = 5.95 eV$

$13.6 Z^{2} (\frac{1}{3^{2}} - \frac{1}{4^{2}}) = 5.95$

$13.6 Z^{2} (\frac{7}{9 \times 16}) = 5.95$

$\Rightarrow Z^{2} = \frac{5.95 \times 9 \times 16}{13.6 \times 7} = 9$

$∴$ $Z = 3$

Q 13 :

Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the $n = 2$ to $n = 1$ transition has energy 74.8 eV higher than the photon emitted in the $n = 3$ to $n = 2$ transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of Z is ______. [2018]

(3)

According to the question, the photon emitted in the $n = 2 \to n = 1$ transition has energy 74.8 eV higher than the photon emitted in the $n = 3 \to n = 2$ transition.

$∴$ $Δ E_{2 \to 1} = 74.8 + Δ E_{3 \to 2}$

$13.6 Z^{2} (1 - \frac{1}{4}) = 74.8 + 13.6 Z^{2} (\frac{1}{4} - \frac{1}{9})$

$∴$ $Z = 3$

Q 14 :

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number $n_{i}$ to another with quantum number $n_{f}$ . $V_{i}$ and $V_{f}$ are respectively the initial and final potential energies of the electron. If $\frac{V_{i}}{V_{f}} = 6.25,$ then the smallest possible $n_{f}$ is [2017]

(5)

Here, $\frac{V_{i}}{V_{f}} = \frac{\frac{- 27.2}{n_{f}^{2}}}{\frac{- 27.2}{n_{i}^{2}}} = \frac{n_{f}^{2}}{n_{i}^{2}} = 6.25$

$∴$ $\frac{n_{f}}{n_{i}} = 2.5 = \frac{5}{2}$

$∴$ Smallest possible $n_{f} = 5$

Q 15 :

A hydrogen atom in its ground state is irradiated by light of wavelength $970 Å$ . Taking $\frac{h c}{e} = 1.237 \times 10^{- 6} eV m$ and the ground state energy of hydrogen atom as $- 13.6 eV$ , the number of lines present in the emission spectrum is [2016]

(6)

Energy of incident light,

$E = \frac{h c}{λ} = \frac{1.237 \times 10^{- 6}}{970 \times 10^{- 10}} eV = 12.75 eV$

$∴$ The energy of electron after absorbing this photon $= - 13.6 + 12.75 = - 0.85 eV$

Let the electron jump to the $n^{th}$ state after excitation.

$∴$ $\frac{- 13.6}{n^{2}} = - 0.85$ $∴$ $n = 4$

$∴$ Total number of spectral lines $= \frac{n (n - 1)}{2} = \frac{4 (4 - 1)}{2} = 6$

Q 16 :

An electron in an excited state of ${Li}^{2 +}$ ion has angular momentum $\frac{3 h}{2 π}$ . The de Broglie wavelength of the electron in this state is $p π a_{0}$ (where $a_{0}$ is the Bohr radius). The value of $p$ is [2015]

(2)

Angular momentum, $L = m v r = \frac{n h}{2 π} = \frac{3 h}{2 π}$ $\Rightarrow n = 3$

$∵$ $λ = \frac{h}{p} = \frac{h}{m v} = \frac{h r}{m v r} \Rightarrow m v r = \frac{h r}{λ}$ $∴$ $\frac{h r}{λ} = \frac{3 h}{2 π}$

$∴$ $λ = \frac{2 π r}{3} = \frac{2}{3} π [a_{0} \frac{n^{2}}{z}]$ $[∵ r = a_{0} \frac{n^{2}}{z}]$

$∴$ $λ = \frac{2}{3} π a_{0} [\frac{3 \times 3}{3}] = 2 π a_{0}$ $[∵ n = 3, z = 3]$

$∴$ $p = 2$

Q 17 :

Consider a hydrogen atom with its electron in the $n^{t h}$ orbital. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of $n$ is $(h c = 1242 e V n m)$ [2015]

(2)

Using the energy conservation principle,

$E_{photon} = E_{ionize} + E_{k}$ $\Rightarrow \frac{h c}{λ} = \frac{13.6}{n^{2}} + E_{k}$

$∴$ $\frac{1242}{90} = \frac{13.6}{n^{2}} + 10.2$ $\Rightarrow n^{2} = 4$

$∴$ $n = 2$

Q 18 :

A particle of mass $m$ is moving in a circular orbit under the influence of the central force $F (r) = - k r,$ corresponding to the potential energy $V (r) = \frac{k r^{2}}{2},$ where $k$ is a positive force constant and $r$ is the radial distance from the origin. According to the Bohr's quantization rule, the angular momentum of the particle is given by $L = n ℏ,$ where $ℏ = \frac{h}{(2 π)},$ $h$ is the Planck's constant, and $n$ a positive integer. If $v$ and $E$ are the speed and total energy of the particle, respectively, then which of the following expression(s) is(are) correct? [2024]

$r^{2} = n ℏ \sqrt{\frac{1}{m k}}$
$v^{2} = n ℏ \sqrt{\frac{k}{m^{3}}}$
$\frac{L}{m r^{2}} = \sqrt{\frac{k}{m}}$
$E = \frac{n ℏ}{2} \sqrt{\frac{k}{m}}$

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Select one or more options

(1, 2, 3)

Mass $m$ is moving in a circular orbit under the influence of the central force $F (r) = - k r$

which provides the necessary centripetal force.

$∴$ $F (r) = F (c)$ or, $k r = \frac{m v^{2}}{r}$

or, $k r^{2} = m v^{2}$ $\dots (i)$

Using the quantisation rule, $n ℏ = m v r$

or, $\frac{n ℏ}{r} = m v$ $\dots (i i)$

Squaring both sides of Eq. (ii), $\frac{n^{2} ℏ^{2}}{r^{2}} = m^{2} v^{2}$ $\dots (i i i)$

Dividing Eq. (i) by Eq. (iii),

$\frac{k r^{2}}{\frac{n^{2} ℏ^{2}}{r^{2}}} = \frac{m v^{2}}{m^{2} v^{2}}$

$\Rightarrow \frac{k}{n^{2} ℏ^{2}} r^{4} = \frac{1}{m}$

$\Rightarrow r = {(\frac{n^{2} ℏ^{2}}{k m})}^{\frac{1}{4}}$ $\Rightarrow r^{2} = n ℏ \sqrt{\frac{1}{m k}}$

Hence, option (1) is correct.

Now using Eq. (i),

$k \cdot \frac{n ℏ}{\sqrt{m k}} = m v^{2}$ $\Rightarrow v^{2} = n ℏ \sqrt{\frac{k}{m^{3}}}$

Hence, option (2) is correct.

Again using Eq. (i),

$\frac{L}{m r^{2}} = \frac{m v r}{m r^{2}} = \frac{v}{r} = \sqrt{\frac{k}{m}}$

Hence, option (3) is correct.

Total energy, $E = K + E$ or,

$E = \frac{1}{2} m v^{2} + \frac{1}{2} k r^{2}$ $= \frac{n ℏ}{2} \sqrt{\frac{k}{m}} + \frac{1}{2} k \cdot \frac{n ℏ}{\sqrt{m k}}$

$∴$ $E = n ℏ \sqrt{\frac{k}{m}}$

Hence, option (4) is incorrect.

Q 19 :

Which of the following statement(s) is (are) correct about the spectrum of hydrogen atom? [2021]

The ratio of the longest wavelength to the shortest wavelength in Balmer series is $\frac{9}{5}$ .
There is an overlap between the wavelength ranges of Balmer and Paschen series.
The wavelengths of Lyman series are given by $(1 + \frac{1}{m^{2}}) λ_{0},$ where $λ_{0}$ is the shortest wavelength of Lyman series and $m$ is an integer.
The wavelength ranges of Lyman and Balmer series do not overlap.

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Select one or more options

(1, 4)

From formula,

(1) $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

For Balmer series, $n_{1} = 2$

$∴$ For longest wavelength, transition occurs from $n = 3$ to $n = 2$

$∴$ $\frac{1}{λ_{\max}} = R [\frac{1}{2^{2}} - \frac{1}{3^{2}}]$

For shortest wavelength, transition occurs from $n = \infty$ to $n = 2$

$∴$ $\frac{1}{λ_{\min}} = R [\frac{1}{2^{2}} - \frac{1}{\infty^{2}}]$

$∴$ $\frac{λ_{longest}}{λ_{shortest}} = \frac{9}{5}$

(2) For Paschen series, $n_{1} = 3$

$λ_{longest of Balmer} = \frac{36}{5 R}$ and

$λ_{shortest of Paschen} = \frac{9}{R}$

Hence there is no overlap between the wavelength ranges of Balmer and Paschen series.

(3), (4) For Lyman series, $n_{1} = 1$

$\frac{1}{λ} = R [1 - \frac{1}{m^{2}}]$

Also, $\frac{1}{λ_{0}} = R$

$∴$ $\frac{1}{λ} = \frac{1}{λ_{0}} [1 - \frac{1}{m^{2}}]$ $\Rightarrow λ = \frac{λ_{0}}{1 - \frac{1}{m^{2}}}$

$λ_{longest of Lyman} = \frac{4}{3 R}$ and $λ_{shortest of Balmer} = \frac{4}{R}$

Hence the wavelength ranges of Lyman and Balmer series do not overlap.

Q 20 :

A particle of mass $m$ moves in circular orbits with potential energy $V (r) = F r,$ where $F$ is a positive constant and $r$ is its distance from the origin. Its energies are calculated using the Bohr model. If the radius of the particle's orbit is denoted by $R$ and its speed and energy are denoted by $v$ and $E$ , respectively, then for the $n^{th}$ orbit (here $h$ is the Planck's constant) [2020]

$R \propto n^{1 / 3}$ and $v \propto n^{2 / 3}$
$R \propto n^{2 / 3}$ and $v \propto n^{1 / 3}$
$E = \frac{3}{2} {(\frac{n^{2} h^{2} F^{2}}{4 π^{2} m})}^{1 / 3}$
$E = 2 {(\frac{n^{2} h^{2} F^{2}}{4 π^{2} m})}^{1 / 3}$

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Select one or more options

(2, 3)

Given: Potential energy of the particle of mass $' m'$ moving in a circular orbit, $V (r) = F r$

$\Rightarrow F = \frac{V (r)}{r}$

Also, centripetal force,

$F = \frac{m v^{2}}{R}$ $\dots (1)$

According to Bohr's second postulate,

$m v R = \frac{n h}{2 π}$ $\Rightarrow v = \frac{n h}{2 π m R}$

Putting this value of $v$ in Eq. (1),

$F = \frac{m}{R} \times \frac{n^{2} h^{2}}{2 π^{2}} \times \frac{1}{m^{2} R^{2}}$ $\Rightarrow R = {(\frac{n^{2} h^{2}}{4 π^{2} m F})}^{1 / 3}$ $∴$ $R \propto n^{2 / 3}$

Now putting this value of $R$ in $v = \frac{n h}{2 π m R}$

$v = \frac{n h}{2 π m} {(\frac{4 π^{2} m F}{n^{2} h^{2}})}^{1 / 3}$ $∴$ $v \propto n^{1 / 3}$

Hence, option (2) is correct.

Total energy, $E = \frac{1}{2} m v^{2} + V (r) = \frac{1}{2} m v^{2} + F R$

$\Rightarrow E = \frac{1}{2} m (\frac{n^{2 / 3} h^{2 / 3} F^{2 / 3}}{2^{2 / 3} π^{2 / 3} m^{4 / 3}}) + F {(\frac{n^{2} h^{2}}{4 π^{2} m F})}^{1 / 3}$

$\Rightarrow E = {(\frac{n^{2} h^{2} F^{2}}{4 π^{2} m})}^{1 / 3} [\frac{1}{2} + 1]$ or, $E = \frac{3}{2} {(\frac{n^{2} h^{2} F^{2}}{4 π^{2} m})}^{1 / 3}$

Hence, option (3) is correct.

Q 21 :

A free hydrogen atom after absorbing a photon of wavelength $λ_{a}$ gets excited from the state $n = 1$ to the state $n = 4$ . Immediately after that the electron jumps to $n = m$ state by emitting a photon of wavelength $λ_{e}$ . Let the change in momentum of atom due to the absorption and the emission be $Δ p_{a}$ and $Δ p_{e}$ , respectively. If $\frac{λ_{a}}{λ_{e}} = \frac{1}{5},$ which of the option(s) is/are correct?

[Use $h c = 1242 eV nm$ ; $1 nm = 10^{- 9} m$ , $h$ and $c$ are Planck's constant and speed of light, respectively] [2019]

$Δ p_{a} / Δ p_{e} = \frac{1}{2}$
The ratio of kinetic energy of the electron in the state $n = m$ to the state $n = 1$ is $\frac{1}{4}$
$m = 2$
$λ_{e} = 418 nm$

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Select one or more options

(2, 3)

(1) Change in linear momentum due to absorption,

$Δ P_{a} = \frac{h}{λ_{a}}$ $(∵ λ = \frac{h}{P})$

Change in linear momentum due to emission,

$Δ P_{e} = \frac{h}{λ_{e}}$ $∵$ $\frac{Δ P_{a}}{Δ P_{e}} = \frac{λ_{e}}{λ_{a}} = 5$ $(∵ \frac{λ_{a}}{λ_{e}} = \frac{1}{5} given)$

So, option (1) is wrong.

(2) Kinetic energy, $K_{n} \propto \frac{Z^{2}}{n^{2}}$ $∴$ $\frac{K_{2}}{K_{1}} = \frac{1^{2}}{2^{2}} = \frac{1}{4}$

So, (2) is correct.

(3) For absorption of energy, $n = 1$ to $n = 4$

$E_{4} - E_{1} = \frac{h c}{λ_{a}} = 13.6 (\frac{1}{1^{2}} - \frac{1}{4^{2}})$ $\dots (i)$

Per emission of energy, $n = m$ to $n = 4$

$E_{4} - E_{m} = \frac{h c}{λ_{e}} = 13.6 (\frac{1}{m^{2}} - \frac{1}{4^{2}})$ $\dots (i i)$

Dividing Eq. (ii) by Eq. (i),

$\frac{λ_{a}}{λ_{e}} = \frac{\frac{1}{m^{2}} - \frac{1}{16}}{1 - \frac{1}{16}} = \frac{1}{5}$

$\Rightarrow \frac{1}{m^{2}} - \frac{1}{16} = \frac{1}{5} \times \frac{15}{16}$ $\Rightarrow \frac{1}{m^{2}} = \frac{3}{16} + \frac{1}{16} = \frac{4}{16}$

$∴$ $m = 2$

So, (3) is correct.

(4) Now from Eq. (ii),

$\frac{h c}{λ_{e}} = 13.6 (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = 13.6 (\frac{1}{4} - \frac{1}{16})$

$∴$ $λ_{e} = \frac{1242 \times 16}{13.6 \times 3} \approx 486.2 nm$

So, option (4) is wrong.

Q 22 :

Highly excited states for hydrogen-like atoms (also called Rydberg states) with nuclear charge Ze are defined by their principal quantum number n, where n >> 1. Which of the following statement(s) is(are) true? [2016]

Relative change in the radii of two consecutive orbitals does not depend on Z
Relative change in the radii of two consecutive orbitals varies as 1/n
Relative change in the energy of two consecutive orbitals varies as $1 / n^{3}$
Relative change in the angular momentum of two consecutive orbitals varies as 1/n

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Select one or more options

(1, 2, 4)

We know radius, $r_{n} = r_{0} \frac{n^{2}}{Z},$

Energy $E_{n} = - \frac{13.6 Z^{2}}{n^{2}},$ and angular momentum, $L_{n} = \frac{n h}{2 π}$

Relative change in the radii of two consecutive orbitals,

$\frac{Δ r}{r} = \frac{r_{n} - r_{n - 1}}{r_{n}} = 1 - \frac{r_{n - 1}}{r_{n}} = 1 - \frac{{(n - 1)}^{2}}{n^{2}}$

$= \frac{2 n - 1}{n^{2}} \approx \frac{2}{n}$ $(∵ n ≫ 1),$ which is independent of $Z$ .

Relative change in the energy of two consecutive orbitals,

$\frac{Δ E}{E} = \frac{E_{n} - E_{n - 1}}{E_{n}} = 1 - \frac{E_{n - 1}}{E_{n}} = 1 - \frac{n^{2}}{{(n - 1)}^{2}}$ $= \frac{- 2 n + 1}{{(n - 1)}^{2}} \approx - \frac{2}{n}$

$\frac{Δ L}{L} = \frac{L_{n} - L_{n - 1}}{L_{n}} = 1 - \frac{L_{n - 1}}{L_{n}} = 1 - \frac{(n - 1)}{n} = \frac{1}{n}$

Q 23 :

The radius of the orbit of an electron in a Hydrogen-like atom is $4.5 a_{0}$ , where $a_{0}$ is the Bohr radius. Its orbital angular momentum is $\frac{3 h}{2 π} .$ It is given that $h$ is Planck's constant and $R$ is the Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are) [2013]

$\frac{9}{32 R}$
$\frac{9}{16 R}$
$\frac{9}{5 R}$
$\frac{4}{3 R}$

1

2

3

4

Select one or more options

(1, 3)

According to Bohr's quantisation, angular momentum

$L = \frac{n h}{2 π} = \frac{3 h}{2 π}$ $∴$ $n = 3$

Also, $r_{n} = \frac{a_{0} n^{2}}{Z} = 4.5 a_{0}$

$∴$ $\frac{n^{2}}{Z} = 4.5$ $\Rightarrow \frac{9}{Z} = 4.5$ $\Rightarrow Z = 2$

According to Rydberg formula,

$\frac{1}{λ} = R Z^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] = 4 R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

For $n_{2} = 3, \to n_{1} = 1$ , $λ = \frac{9}{8 \times 4 R} = \frac{9}{32 R}$

For $n_{2} = 3, \to n_{1} = 2$ , $λ = \frac{36}{5 \times 4 R} = \frac{9}{5 R}$

For $n_{2} = 2, \to n_{1} = 1$ , $λ = \frac{4}{3 \times 4 R} = \frac{1}{3 R}$

Q 24 :

Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in the ORS. [2007]

Column I Column II

(A) Transition between two atomic energy levels (p) Characteristic X-rays

(B) Electron emission from a material (q) Photoelectric effect

(C) Moseley's law (r) Hydrogen spectrum

(D) Change of photon energy into kinetic energy of electrons (s) $β$ -decay

$A \to q; B \to p, r; C \to q, s; D \to p$
$A \to q; B \to p; C \to q, s; D \to p, r$
$A \to q; B \to q, s; C \to p; D \to p, r$
$A \to p, r; B \to q, s; C \to p; D \to q$

1

2

3

4

(4)

$A \to p, r$

The lines in the hydrogen spectrum are obtained due to the transition of electrons from one energy level to another. Characteristic X-rays are produced due to the transition of electrons from one energy level to another.

$B \to q, s$

In the photoelectric effect, electrons are emitted from the metal surface when light of appropriate frequency is incident on it.

In $β$ -decay, electrons are emitted from the nucleus of an atom.

$C \to p$

According to Moseley's law, the frequency of emitted X-rays is related to the atomic number $(z)$ of the target material as $\sqrt{ν} = a (Z - b)$

$D \to q$

According to Einstein's photoelectric equation, the energy of photons of the incident radiation is converted into the kinetic energy of emitted electrons. $K . E_{\max} = h ν - ϕ$

Q 25 :

The key feature of Bohr's theory of the spectrum of the hydrogen atom is the quantization of angular momentum when an electron revolves around a proton. We will extend this to a general rotational motion to find the quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition. [2010]

Q. A diatomic molecule has moment of inertia $I$ . By Bohr's quantization condition its rotational energy in the $n^{th}$ level ( $n = 0$ is not allowed) is

$\frac{1}{n^{2}} (\frac{h^{2}}{8 π^{2} I})$
$\frac{1}{n} (\frac{h^{2}}{8 π^{2} I})$
$n (\frac{h^{2}}{8 π^{2} I})$
$n^{2} (\frac{h^{2}}{8 π^{2} I})$

1

2

3

4

(4)

Rotational kinetic energy, $K . E ._{rot} = \frac{1}{2} I ω^{2} = \frac{1}{2} I {(\frac{L}{I})}^{2}$ $[∵ L = I ω]$

And according to Bohr's quantisation principle, $L = \frac{n h}{2 π}$

$∴$ $K . E ._{rot} = \frac{1}{2} \frac{L^{2}}{I} = \frac{1}{2 I} \times \frac{n^{2} h^{2}}{4 π^{2}} = n^{2} (\frac{h^{2}}{8 π^{2} I})$

Q 26 :

The key feature of Bohr's theory of the spectrum of the hydrogen atom is the quantization of angular momentum when an electron revolves around a proton. We will extend this to a general rotational motion to find the quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition. [2010]

Q. It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close to $\frac{4}{π} \times 10^{11} Hz .$ Then the moment of inertia of the CO molecule about its center of mass is close to (Take $h = 2 π \times 10^{- 34} J s$ )

$2.76 \times 10^{- 46} {kg m}^{2}$
$1.87 \times 10^{- 46} {kg m}^{2}$
$4.67 \times 10^{- 47} {kg m}^{2}$
$1.17 \times 10^{- 47} {kg m}^{2}$

1

2

3

4

(2)

$Rotational kinetic energy K . E ._{rot} = \frac{1}{2} I ω^{2} = \frac{1}{2} I {(\frac{L}{I})}^{2}$ $[∵ L = I ω]$

And according to Bohr's quantisation principle, $L = \frac{n h}{2 π}$

$∴$ $K . E ._{rot} = \frac{1}{2} \frac{L^{2}}{I} = \frac{1}{2 I} \times \frac{n^{2} h^{2}}{4 π^{2}} = n^{2} [\frac{h^{2}}{8 π^{2} I}]$ $\dots (i)$

From ground state $(n = 1)$ to first excited state $(n = 2)$ ,

Energy given = change in kinetic energy

$h ν = K_{f} - K_{i} = \frac{h^{2}}{8 π^{2} I} (2^{2} - 1^{2})$ $[From Eq. (i)]$

$h ν = \frac{3 h^{2}}{8 π^{2} I}$ $\Rightarrow I = \frac{3 h}{8 π^{2} ν} = \frac{3 \times 2 π \times 10^{- 34}}{8 π^{2} \times \frac{4}{π} \times 10^{11}} = \frac{3}{16} \times 10^{- 45}$

$I = 1.87 \times 10^{- 46} {kg m}^{2}$

Q 27 :

The key feature of Bohr's theory of the spectrum of the hydrogen atom is the quantization of angular momentum when an electron revolves around a proton. We will extend this to a general rotational motion to find the quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition. [2010]

Q. In a CO molecule, the distance between C (mass = 12 a.m.u.) and O (mass = 16 a.m.u.), where 1 a.m.u. $= \frac{5}{3} \times 10^{- 27} kg$ , is close to

$2.4 \times 10^{- 10} m$
$1.9 \times 10^{- 10} m$
$1.3 \times 10^{- 10} m$
$4.4 \times 10^{- 11} m$

1

2

3

4

(3)

Moment of inertia of CO molecule,

$I = μ r^{2}$ $\Rightarrow r^{2} = \frac{I}{μ}$

where, $μ$ = reduced mass of the CO molecule and $r$ = distance between C and O

Reduced mass of the CO molecule,

$μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}} = [\frac{(12) (16)}{12 + 16}] \times \frac{5}{3} \times 10^{- 27} kg$

But, $I = 1.87 \times 10^{- 46} {kg m}^{2}$ (from the above question)

$∴$ $r^{2} = [\frac{1.87 \times 10^{- 46}}{\frac{12 \times 16}{28} \times \frac{5}{3} \times 10^{- 27}}] = \frac{1.87 \times 10^{- 46} \times 28 \times 3}{12 \times 16 \times 5 \times 10^{- 27}}$

$∴$ $r = 1.3 \times 10^{- 10} m$

Q 28 :

When a particle is restricted to move along the $x$ -axis between $x = 0$ and $x = a$ , where $a$ is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends $x = 0$ and $x = a$ . The wavelength of this standing wave is related to the linear momentum $p$ of the particle according to the de Broglie relation. The energy of the particle of mass $m$ is related to its linear momentum as $E = \frac{p^{2}}{2 m} .$ Thus, the energy of the particle can be denoted by a quantum number $' n'$ taking values $1, 2, 3, \dots$ ( $n = 1$ , called the ground state) corresponding to the number of loops in the standing wave.

Use the model described above to answer the following three questions for a particle moving in the line $x = 0$ to $x = a$ . Take $h = 6.6 \times 10^{- 34} J s$ and $e = 1.6 \times 10^{- 19} C$ .

Q. The allowed energy for the particle for a particular value of $n$ is proportional to [2009]

$a^{- 2}$
$a^{- 3 / 2}$
$a^{- 1}$
$a^{2}$

1

2

3

4

(1)

Energy, $E = \frac{h^{2}}{2 m λ^{2}}$ $(∵ E = \frac{P^{2}}{2 m} and P = \frac{h}{λ})$

The length in which the particle is restricted to move is

$a = \frac{n λ}{2}$ $\Rightarrow λ = \frac{2 a}{n}$

Putting this value of $λ$ , we get

$E = \frac{h^{2} n^{2}}{2 m \times 4 a^{2}} = \frac{n^{2} h^{2}}{8 m a^{2}}$

$∴$ $E \propto a^{- 2}$

Q 29 :

When a particle is restricted to move along the $x$ -axis between $x = 0$ and $x = a$ , where $a$ is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends $x = 0$ and $x = a$ . The wavelength of this standing wave is related to the linear momentum $p$ of the particle according to the de Broglie relation. The energy of the particle of mass $m$ is related to its linear momentum as $E = \frac{p^{2}}{2 m} .$ Thus, the energy of the particle can be denoted by a quantum number $n$ taking values $1, 2, 3, \dots$ ( $n = 1$ , called the ground state) corresponding to the number of loops in the standing wave.

Use the model described above to answer the following three questions for a particle moving in the line $x = 0$ to $x = a$ . Take $h = 6.6 \times 10^{- 34} J s$ and $e = 1.6 \times 10^{- 19} C$ .

Q. If the mass of the particle is $m = 1.0 \times 10^{- 30} kg$ and $a = 6.6 nm$ the energy of the particle in its ground state is closest to [2009]

0.8 meV
8 meV
80 meV
800 meV

1

2

3

4

(2)

For ground state, $n = 1$

Given, $m = 1.0 \times 10^{- 30} kg,$ $a = 6.6 \times 10^{- 9} m$

$∴$ $E = \frac{n^{2} h^{2}}{8 m a^{2}} = \frac{1^{2} \times {(6.6 \times 10^{- 34})}^{2}}{8 \times 1 \times 10^{- 30} \times {(6.6 \times 10^{- 9})}^{2}} J$

$= 8 meV$

Q 30 :

When a particle is restricted to move along the $x$ -axis between $x = 0$ and $x = a$ , where $a$ is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends $x = 0$ and $x = a$ . The wavelength of this standing wave is related to the linear momentum $p$ of the particle according to the de Broglie relation. The energy of the particle of mass $m$ is related to its linear momentum as $E = \frac{p^{2}}{2 m} .$ Thus, the energy of the particle can be denoted by a quantum number $' n'$ taking values $1, 2, 3, \dots$ ( $n = 1$ , called the ground state) corresponding to the number of loops in the standing wave.

Use the model described above to answer the following three questions for a particle moving in the line $x = 0$ to $x = a$ . Take $h = 6.6 \times 10^{- 34} J s$ and $e = 1.6 \times 10^{- 19} C.$

Q. The speed of the particle, that can take discrete values, is proportional to [2009]

$n^{- 3 / 2}$
$n^{- 1}$
$n^{1 / 2}$
$n$

1

2

3

4

(4)

$λ = \frac{h}{p}$ $\Rightarrow λ = \frac{h}{m v}$

But, $\frac{n λ}{2} = a$ $\Rightarrow λ = \frac{2 a}{n}$

$∴$ $\frac{h}{m v} = \frac{2 a}{n}$

$∴$ $m v = \frac{n h}{2 a}$ $\Rightarrow v = \frac{n h}{2 a m}$

$∴$ $v \propto n$

Topic Question Set

Q 2 :

In a hydrogen-like atom electron make transition from an energy level with quantum number $n$ to another with quantum number $(n - 1)$ . If $n ≫ 1$ , the frequency of radiation emitted is proportional to: [2012]

Q 3 :

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 $Å$ . The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is [2011]

Q 4 :

The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is [2007]

Q 6 :

If the atom ${}_{100}{Fm}^{257}$ follows the Bohr model and the radius of ${}_{100}{Fm}^{257}$ is $n$ times the Bohr radius, then find $n$ . [2003]

Q 7 :

The electric potential between a proton and an electron is given by $V = V_{0} \ln (\frac{r}{r_{0}}),$ where $r_{0}$ is a constant. Assuming Bohr's model to be applicable, write variation of $r_{n}$ with $n$ , $n$ being the principal quantum number? [2003]

Q 8 :

A Hydrogen atom and a ${Li}^{+ +}$ ion are both in the second excited state. If $ℓ_{H}$ and $ℓ_{L i}$ are their respective electronic angular momenta, and $E_{H}$ and $E_{L i}$ their respective energies, then [2002]

Q 9 :

The transition from the state $n = 4$ to $n = 3$ in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition. [2001]

Q 10 :

The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true? [2000]

Q 15 :

A hydrogen atom in its ground state is irradiated by light of wavelength $970 Å$ . Taking $\frac{h c}{e} = 1.237 \times 10^{- 6} eV m$ and the ground state energy of hydrogen atom as $- 13.6 eV$ , the number of lines present in the emission spectrum is [2016]

Q 16 :

An electron in an excited state of ${Li}^{2 +}$ ion has angular momentum $\frac{3 h}{2 π}$ . The de Broglie wavelength of the electron in this state is $p π a_{0}$ (where $a_{0}$ is the Bohr radius). The value of $p$ is [2015]

Q 17 :

Consider a hydrogen atom with its electron in the $n^{t h}$ orbital. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of $n$ is $(h c = 1242 e V n m)$ [2015]

Q 19 :

Which of the following statement(s) is (are) correct about the spectrum of hydrogen atom? [2021]

Q 22 :

Highly excited states for hydrogen-like atoms (also called Rydberg states) with nuclear charge Ze are defined by their principal quantum number n, where n >> 1. Which of the following statement(s) is(are) true? [2016]

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	Column I		Column II
(A)	Transition between two atomic energy levels	(p)	Characteristic X-rays
(B)	Electron emission from a material	(q)	Photoelectric effect
(C)	Moseley's law	(r)	Hydrogen spectrum
(D)	Change of photon energy into kinetic energy of electrons	(s)	$β$ -decay

Topic Question Set

Q 2 : In a hydrogen-like atom electron make transition from an energy level with quantum number n to another with quantum number (n-1). If n≫1, the frequency of radiation emitted is proportional to: [2012]

Q 3 : The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is [2011]

Q 4 : The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is [2007]

Q 6 : If the atom Fm100257 follows the Bohr model and the radius of Fm100257 is n times the Bohr radius, then find n. [2003]

Q 7 : The electric potential between a proton and an electron is given by V=V0ln(rr0), where r0 is a constant. Assuming Bohr's model to be applicable, write variation of rn with n, n being the principal quantum number? [2003]

Q 8 : A Hydrogen atom and a Li++ ion are both in the second excited state. If ℓH and ℓLi are their respective electronic angular momenta, and EH and ELi their respective energies, then [2002]

Q 9 : The transition from the state n=4 to n=3 in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition. [2001]

Q 10 : The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true? [2000]

Q 14 : An electron in a hydrogen atom undergoes a transition from an orbit with quantum number ni to another with quantum number nf. Vi and Vf are respectively the initial and final potential energies of the electron. If ViVf=6.25, then the smallest possible nf is [2017]

Q 15 : A hydrogen atom in its ground state is irradiated by light of wavelength 970 Å. Taking hce=1.237×10-6 eV m and the ground state energy of hydrogen atom as -13.6 eV, the number of lines present in the emission spectrum is [2016]

Q 16 : An electron in an excited state of Li2+ ion has angular momentum 3h2π. The de Broglie wavelength of the electron in this state is pπa0 (where a0 is the Bohr radius). The value of p is [2015]

Q 17 : Consider a hydrogen atom with its electron in the nth orbital. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of n is (hc=1242 eV nm) [2015]

Q 19 : Which of the following statement(s) is (are) correct about the spectrum of hydrogen atom? [2021]

Q 22 : Highly excited states for hydrogen-like atoms (also called Rydberg states) with nuclear charge Ze are defined by their principal quantum number n, where n >> 1. Which of the following statement(s) is(are) true? [2016]

Q 23 : The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a0, where a0 is the Bohr radius. Its orbital angular momentum is 3h2π. It is given that h is Planck's constant and R is the Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are) [2013]

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Q 2 :

In a hydrogen-like atom electron make transition from an energy level with quantum number $n$ to another with quantum number $(n - 1)$ . If $n ≫ 1$ , the frequency of radiation emitted is proportional to: [2012]

Q 3 :

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 $Å$ . The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is [2011]

Q 4 :

The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is [2007]

Q 6 :

If the atom ${}_{100}{Fm}^{257}$ follows the Bohr model and the radius of ${}_{100}{Fm}^{257}$ is $n$ times the Bohr radius, then find $n$ . [2003]

Q 7 :

The electric potential between a proton and an electron is given by $V = V_{0} \ln (\frac{r}{r_{0}}),$ where $r_{0}$ is a constant. Assuming Bohr's model to be applicable, write variation of $r_{n}$ with $n$ , $n$ being the principal quantum number? [2003]

Q 8 :

A Hydrogen atom and a ${Li}^{+ +}$ ion are both in the second excited state. If $ℓ_{H}$ and $ℓ_{L i}$ are their respective electronic angular momenta, and $E_{H}$ and $E_{L i}$ their respective energies, then [2002]

Q 9 :

The transition from the state $n = 4$ to $n = 3$ in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition. [2001]

Q 10 :

The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true? [2000]

Q 15 :

A hydrogen atom in its ground state is irradiated by light of wavelength $970 Å$ . Taking $\frac{h c}{e} = 1.237 \times 10^{- 6} eV m$ and the ground state energy of hydrogen atom as $- 13.6 eV$ , the number of lines present in the emission spectrum is [2016]

Q 16 :

An electron in an excited state of ${Li}^{2 +}$ ion has angular momentum $\frac{3 h}{2 π}$ . The de Broglie wavelength of the electron in this state is $p π a_{0}$ (where $a_{0}$ is the Bohr radius). The value of $p$ is [2015]

Q 17 :

Consider a hydrogen atom with its electron in the $n^{t h}$ orbital. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of $n$ is $(h c = 1242 e V n m)$ [2015]

Q 19 :

Which of the following statement(s) is (are) correct about the spectrum of hydrogen atom? [2021]

Q 22 :

Highly excited states for hydrogen-like atoms (also called Rydberg states) with nuclear charge Ze are defined by their principal quantum number n, where n >> 1. Which of the following statement(s) is(are) true? [2016]