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 $Z = 100$

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

$∴$ $n = \frac{1}{4}$

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]

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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]

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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]