Atomic Structure | Physics JEE previous year questions with Solutions

Q 1 :

The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is [2024]

2 : 1
1 : 4
1 : 2
4 : 1

1

2

3

4

(4)

$For Balmer series limit, \frac{1}{λ_{B}} = R [\frac{1}{{(2)}^{2}} - \frac{1}{\infty^{2}}] = \frac{R}{4}$

$For Lyman series limit, \frac{1}{λ_{L}} = R [\frac{1}{{(1)}^{2}} - \frac{1}{\infty^{2}}] = R$

$\Rightarrow \frac{λ_{B}}{λ_{L}} = 4 : 1$

Q 2 :

The longest wavelength associated with Paschen series is

$(Given R_{H} = 1.097 \times 10^{7} in SI unit)$ [2024]

$2.973 \times 10^{- 6} m$
$3.646 \times 10^{- 6} m$
$1.094 \times 10^{- 6} m$
$1.876 \times 10^{- 6} m$

1

2

3

4

(4)

For longest wavelength in Paschen's series:

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

$n_{1} = 3 and n_{2} = 4$

$\frac{1}{λ} = R [\frac{1}{{(3)}^{2}} - \frac{1}{{(4)}^{2}}] = R [\frac{16 - 9}{16 \times 9}]$

$\Rightarrow λ = \frac{16 \times 9}{7 R} = \frac{16 \times 9}{7 \times 1.097 \times 10^{7}} = 1.876 \times 10^{- 6} m$

Q 3 :

If the wavelength of the first member of Lyman series of hydrogen is $λ$ . The wavelength of the second member will be [2024]

$\frac{5}{27} λ$
$\frac{27}{32} λ$
$\frac{32}{27} λ$
$\frac{27}{5} λ$

1

2

3

4

(2)

$\frac{1}{λ} = \frac{13.6 z^{2}}{h c} [\frac{1}{1^{2}} - \frac{1}{2^{2}}]$ ...(i)

$\frac{1}{λ^{'}} = \frac{13.6 z^{2}}{h c} [\frac{1}{1^{2}} - \frac{1}{3^{2}}]$ ...(ii)

$O n d i v i d i n g (i) a n d (i i) λ^{'} = \frac{27}{32} λ$

Q 4 :

The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly [2024]

1.5 eV
13.6 eV
1.9 eV
12.1 eV

1

2

3

4

(4)

$Transition from n \to 1 to n \to 3$

$E = 13.6 \times Z [\frac{1}{n_{2}^{2}} - \frac{1}{n_{1}^{2}}]$

$= 13.6 \times 1 \times [\frac{1}{1} - \frac{1}{9}]$

$= 13.6 \times \frac{8}{9} = 12.1 eV$

Q 5 :

The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is 915 $Å$ .The longest wavelength of spectral lines in the Balmar series will be ________ $Å$ . [2024]

(6588) $\frac{h c}{λ} = - 13.6 (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

Lyman series, shortest wavelength, highest energetic

$\frac{h c}{λ_{0}} = - 13.6 \times 1$

Balmer Series, $n_{1} = 2, n_{2} = 3$ (for longest wavelength)

$\frac{h c}{λ_{1}} = - 13.6 (\frac{1}{2^{2}} - \frac{1}{3^{2}})$

$= - 13.6 \times \frac{5}{36}$

$λ_{1} = \frac{36}{5} \cdot λ_{0} = \frac{36}{5} \times 915 Å = 6588 Å$

Q 6 :

Hydrogen atom is bombarded with electrons accelerated through a potential difference of V, which causes excitation of hydrogen atoms. If the experiment is being performed at T = 0 K, the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be $\frac{α}{10} V$ , where $α$ = _______ . [2024]

(121) $Δ E_{\min} = E_{3} - E_{1}$

$= - 1.51 - (- 13.6)$

$Δ E_{\min} = 12.09 eV$

$V_{\min} = 12.09 volt = \frac{α}{10}$

$α = 120.9 = 121$

Q 7 :

A particular hydrogen like ion emits the radiation of frequency $3 \times 10^{15}$ Hz when it makes transition from $n = 2$ to $n = 1$ . The frequency of radiation emitted in transition from $n = 3$ to $n = 1$ is $\frac{x}{9} \times 10^{15}$ Hz, when $x =$ __________ . [2024]

(32) $E = - 13.6 Z^{2} (\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}})$

$E = C (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$

$h ν = C [\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}]$

$\frac{ν_{1}}{ν_{2}} = \frac{{[\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}]}_{2 - 1}}{{[\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}]}_{3 - 1}}$

$= \frac{[\frac{1}{1} - \frac{1}{4}]}{[\frac{1}{1} - \frac{1}{9}]} = \frac{3 / 4}{8 / 9} = \frac{3}{4} \times \frac{9}{8}$

$\frac{ν_{1}}{ν_{2}} = \frac{27}{32}$

$ν_{2} = \frac{32}{27} ν_{1} = \frac{32}{27} \times 3 \times 10^{15} Hz = \frac{32}{9} \times 10^{15} H z$

$Given, f_{2} = \frac{x}{9} \times 10^{15} Hz$

$\Rightarrow x = 32$

Q 8 :

During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is 2000 $\overset{°}{A}$ and it becomes 6000 $\overset{°}{A}$ when the electron jumps from state B to state C. Then the wavelength of the radiation emitted during the transition of electrons from state A to state B is [2025]

3000 $\overset{°}{A}$
6000 $\overset{°}{A}$
4000 $\overset{°}{A}$
2000 $\overset{°}{A}$

1

2

3

4

(1)

For A to C : $E_{A} - E_{C} = \frac{h c}{2000 \overset{°}{A}}$ ... (i)

and for B to C : $E_{B} - E_{C} = \frac{h c}{6000 \overset{°}{A}}$ ... (ii)

Now for A to B : $E_{A} - E_{B} = (E_{A} - E_{C}) - (E_{B} - E_{C})$

$\frac{h c}{λ_{A B}} = \frac{h c}{2000} - \frac{h c}{6000}$

$\frac{1}{λ_{A B}} = \frac{1}{3000 \overset{°}{A}} \Rightarrow λ_{A B} = 3000 \overset{°}{A}$

Q 9 :

The number of spectral lines emitted by atomic hydrogen that is in the $4^{th}$ energy level, is [2025]

6
0
3
1

1

2

3

4

(1)

Number of spectral line (N) = $\frac{n (n - 1)}{2}$ . here n = 4

So, N = 6

Total posible transition = 6

Q 10 :

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is. [2025]

5 : 36
5 : 27
3 : 4
27 : 5

1

2

3

4

(2)

Largest wavelength of Lyman series,

$\frac{1}{λ_{1}} = R [\frac{1}{1} - \frac{1}{4}] = \frac{3 R}{4}$

$λ_{1} = \frac{4}{3 R}$ ... (i)

Largest wavelength of Balmer series

$\frac{1}{λ_{2}} = R [\frac{1}{4} - \frac{1}{9}] = \frac{5 R}{36}$

$λ_{2} = \frac{36}{5 R}$

Then, $\frac{λ_{1}}{λ_{2}} = \frac{5}{27}$

Q 11 :

The energy levels of an atom is shown in figure. Which one of these transitions will result in the emission of a photon of wavelength 124.1 nm? Given ( $h = 6.62 \times 10^{- 34} Js$ ) [2023]

D
B
A
C

1

2

3

4

(1)

$λ = \frac{h c}{Δ E}$

$Δ E_{A} = 2.2 eV$

$Δ E_{B} = 5.2 eV$

$Δ E_{C} = 3 eV$

$Δ E_{D} = 10 eV$

$λ_{A} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{2.2 \times 1.6 \times 10^{- 19}} = \frac{12.41 \times 10^{- 7}}{2.2} m = 564 nm$

$λ_{B} = \frac{1241}{5.2} nm = 238.65 nm$

$λ_{C} = \frac{1241}{3} nm = 413.66 nm$

$λ_{D} = \frac{1241}{10} nm = 124.1 nm$

Q 12 :

An electron of a hydrogen like atom, having Z = 4, jumps in from $4^{th}$ energy state to $2^{nd}$ energy state. The energy released in this process, will be: (Given $R_{c h} = 13.6 eV$ ) Where R = Rydberg constant c = Speed of light in vacuum h = Planck's constant [2023]

13.6 eV
10.5 eV
3.4 eV
40.8 eV

1

2

3

4

(4)

$Δ E = 13.6 Z^{2} [\frac{1}{2^{2}} - \frac{1}{4^{2}}] eV$

$= 13.6 \times {(4)}^{2} (\frac{1}{4} - \frac{1}{16}) eV$

$= 13.6 [4 - 1] eV = 13.6 \times 3 = 40.8 eV$

Q 13 :

The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is [2023]

C
A
D
B

1

2

3

4

(3)

$Δ E = \frac{h c}{λ} \Rightarrow λ \propto \frac{1}{Δ E}$

For shortest wavelength, energy gap should be maximum.

So, correct choice is transition from $n = 3$ to $n = 1$ .

Q 14 :

The angular momentum for the electron in Bohr’s orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be [2023]

$\frac{L}{2}$
$zero$
$L$
$2 L$

1

2

3

4

(3)

$L = m v r, r \propto n^{2}, v \propto \frac{1}{n}$
$∴$ $L \propto n$

Also, $L = \frac{n h}{2 π}$ ,

Bohr orbit is, $L_{1} = L = \frac{1 h}{2 π}$

$L_{2} = 2 [L] = 2 L$

$L_{2} = \frac{2 h}{2 π}$

So change = $L_{2} - L_{1} = 2 L - L = L$

Q 15 :

The energy of ${He}^{+}$ ion in its first excited state is. [2023]

(The ground state energy for the Hydrogen atom is −13.6 eV)

−3.4 eV
−54.4 eV
−13.6 eV
−27.2 eV

1

2

3

4

(3)

$E_{a} = \frac{- 13.6 Z^{2}}{n^{2}} = \frac{- 13.6 \times 4}{4} = - 13.6 eV$

Q 16 :

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be [2023]

2
1
3
4

1

2

3

4

(3)

According to Bohr's postulates, an electron makes jump to higher energy orbital if it absorbs a photon of energy equal to difference between the energies of an excited state and the ground state. Assuming that collided electron takes energy equal to 10.2 eV or 12.09 eV from the incoming electron beam (some part lost due to collision). The maximum excited state is $n = 3$ . So, number of spectral lines $= \frac{3 (3 - 1)}{2} = 3$

Q 17 :

For hydrogen atom, $λ_{1}$ and $λ_{2}$ are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of $λ_{1}$ and $λ_{2}$ is $\frac{x}{32}$ . The value of $x$ is _________. [2023]

(27)

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

$\frac{1}{λ_{1}} = R z^{2} [\frac{1}{1^{2}} - \frac{1}{3^{2}}] = \frac{8}{9} R z^{2} \dots (1)$

$\frac{1}{λ_{2}} = R z^{2} [\frac{1}{1^{2}} - \frac{1}{2^{2}}] = \frac{3}{4} R z^{2} \dots (2)$

$\frac{1}{2} \Rightarrow \frac{λ_{2}}{λ_{1}} = \frac{8}{9} \times \frac{4}{3} = \frac{32}{27} \Rightarrow \frac{λ_{1}}{λ_{2}} = \frac{27}{32}$

Q 18 :

The wavelength of the radiation emitted is $λ_{0}$ when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be $\frac{20}{x} λ_{0}$ . The value of $' x'$ is ________. [2023]

(27)

$\frac{1}{λ_{0}} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

$\frac{1}{λ_{0}} = R [\frac{1}{4} - \frac{1}{9}] = \frac{5 R}{36} \dots (i)$

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

$\frac{1}{2} \Rightarrow \frac{λ}{λ_{0}} = \frac{5}{36} \times \frac{16}{3} = \frac{20}{27}$

Q 19 :

The radius of fifth orbit of the ${Li}^{+ +}$ is _______ $\times 10^{- 12} m$ .

Take: radius of hydrogen atom $= 0.51 Å$ . [2023]

(425)

$r_{n} = r_{0} \frac{n^{2}}{Z} \Rightarrow r_{n} = 0.51 \times \frac{25}{3} Å = 4.25 \times 10^{- 10} m$

Q 20 :

The ratio of wavelength of spectral lines $H_{α}$ and $H_{β}$ in the Balmer series is $\frac{x}{20}$ . The value of $x$ is ________. [2023]

(27)

$\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] for H-atom$

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

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

$For H_{α}, n_{2} = 3 and H_{β}, n_{2} = 4$

$\frac{1}{λ_{H_{α}}} = R [\frac{1}{4} - \frac{1}{9}] = \frac{5 R}{36}$

$\frac{1}{λ_{H_{β}}} = R [\frac{1}{4} - \frac{1}{16}] = \frac{3 R}{16}$

$\frac{\frac{1}{λ_{H_{α}}}}{\frac{1}{λ_{H_{β}}}} = \frac{\frac{5 R}{36}}{\frac{3 R}{16}}$

$\Rightarrow \frac{λ_{H_{α}}}{λ_{H_{β}}} = \frac{27}{20} = \frac{x}{20} \Rightarrow x = 27$

Q 21 :

If 917 $Å$ be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be _________ $Å$ . [2023]

(3668)

$For lowest wavelength of Lyman series, \frac{1}{λ} = R Z^{2} [\frac{1}{1^{2}} - \frac{1}{\infty^{2}}] = R Z^{2}$

$For lowest wavelength of Balmer series, \frac{1}{λ'} = R Z^{2} [\frac{1}{2^{2}} - \frac{1}{\infty^{2}}] = \frac{R Z^{2}}{4}$

$λ' = \frac{4}{R Z^{2}} = 4 \times 917 = 3668 Å$

Q 22 :

A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is $x \times 10^{15} Hz$ . The value of $x$ is _______. (Given $h = 4.25 \times 10^{- 15} eV s$ ) [2023]

(3)

$6 = {}^{4}C_{2} \Rightarrow n_{2} = 4$

$h ν = E_{4} - E_{1}$

$∴$ $ν = 13.6 (\frac{1}{1^{2}} - \frac{1}{4^{2}}) \times \frac{1}{4.25 \times 10^{- 15}} = 3 \times 10^{15} Hz$

Q 23 :

An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 600 nm. The net energy absorbed by the atom in this process is $n \times 10^{- 4} eV$ . The value of $n$ is ________. [2023]

[Assume the atom to be stationary during the absorption and emission process]

(Take $h = 6.6 \times 10^{- 34} Js$ and $c = 3 \times 10^{8} m / s$ )

(4125)

$E = E_{1} - E_{2} = \frac{h c}{λ_{1}} - \frac{h c}{λ_{2}} = h c (\frac{1}{λ_{1}} - \frac{1}{λ_{2}})$

$= 6.6 \times 10^{- 34} \times 3 \times 10^{8} (\frac{1}{500 \times 10^{- 9}} - \frac{1}{600 \times 10^{- 9}})$

$= 6.6 \times 10^{- 20} J$

$= \frac{6.6 \times 10^{- 20}}{1.6 \times 10^{- 19}} eV = 4.125 \times 10^{- 1} eV = 4125 \times 10^{- 4} eV$

Q 24 :

As per given figure A, B and C are the first, second and third excited energy level of hydrogen atom respectively. If the ratio of the two wavelengths (i.e., $\frac{λ_{1}}{λ_{2}}$ ) is $\frac{7}{4 n}$ , then the value of $n$ will be _______. [2023]

(5)

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

$\frac{1}{λ_{1}} = R {(1)}^{2} [\frac{1}{{(2)}^{2}} - \frac{1}{{(3)}^{2}}] = R (\frac{5}{36}) \dots (i)$

$\frac{1}{λ_{2}} = R {(1)}^{2} [\frac{1}{{(3)}^{2}} - \frac{1}{{(4)}^{2}}] = R (\frac{7}{144}) \dots (i i)$

(ii) + (i) gives

$\frac{λ_{1}}{λ_{2}} = \frac{\frac{7}{144}}{\frac{5}{36}} = \frac{7}{20} = \frac{7}{4 \times 5}$

$∴$ $n = 5$

Q 25 :

The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly ______ nm. [2026]

1217
1550
1784
1875

1

2

3

4

(1)

$n = 4$ ___________________

$n = 3$ ___________________ $Paschen$

$n = 2$ ___________________ $Balmer$

$n = 1$ ___________________ $Lyman$

$Smallest wavelength of Lyman$

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

$R = \frac{1}{λ} = \frac{1}{91} {nm}^{- 1}$

$λ_{\max}$ for Balmer series

$n_{1} = 2 \to n_{2} = 3$

$\frac{1}{λ_{B}} = R (\frac{1}{4} - \frac{1}{9})$

$\frac{1}{λ_{B}} = \frac{1}{91} (\frac{5}{36})$

$λ_{B} = (\frac{91 \times 36}{5}) = 655.2 nm$

$λ_{\max}$ Paschen

$n_{1} = 3 \to n_{2} = 4$

$\frac{1}{λ_{p}} = \frac{1}{91} (\frac{1}{3^{2}} - \frac{1}{4^{2}}) = \frac{1}{91} (\frac{7}{144})$

$λ_{p} = (\frac{91 \times 144}{7}) = 1872 nm$

$Δ λ = λ_{p} - λ_{B}$

$Δ λ = 1872 - 655.2$

$Δ λ = 1216.8$

$Δ λ \approx 1217$

Topic Question Set

Q 1 :

The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is [2024]

Q 2 :

The longest wavelength associated with Paschen series is

$(Given R_{H} = 1.097 \times 10^{7} in SI unit)$ [2024]

Q 3 :

If the wavelength of the first member of Lyman series of hydrogen is $λ$ . The wavelength of the second member will be [2024]

Q 4 :

The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly [2024]

Q 5 :

The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is 915 $Å$ .The longest wavelength of spectral lines in the Balmar series will be ________ $Å$ . [2024]

Q 7 :

A particular hydrogen like ion emits the radiation of frequency $3 \times 10^{15}$ Hz when it makes transition from $n = 2$ to $n = 1$ . The frequency of radiation emitted in transition from $n = 3$ to $n = 1$ is $\frac{x}{9} \times 10^{15}$ Hz, when $x =$ __________ . [2024]

Q 9 :

The number of spectral lines emitted by atomic hydrogen that is in the $4^{th}$ energy level, is [2025]

Q 10 :

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is. [2025]

Q 11 :

The energy levels of an atom is shown in figure. Which one of these transitions will result in the emission of a photon of wavelength 124.1 nm? Given ( $h = 6.62 \times 10^{- 34} Js$ ) [2023]

Q 12 :

An electron of a hydrogen like atom, having Z = 4, jumps in from $4^{th}$ energy state to $2^{nd}$ energy state. The energy released in this process, will be: (Given $R_{c h} = 13.6 eV$ ) Where R = Rydberg constant c = Speed of light in vacuum h = Planck's constant [2023]

Q 13 :

The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is [2023]

Q 14 :

The angular momentum for the electron in Bohr’s orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be [2023]

Q 15 :

The energy of ${He}^{+}$ ion in its first excited state is. [2023]

(The ground state energy for the Hydrogen atom is −13.6 eV)

Q 16 :

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be [2023]

Q 17 :

For hydrogen atom, $λ_{1}$ and $λ_{2}$ are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of $λ_{1}$ and $λ_{2}$ is $\frac{x}{32}$ . The value of $x$ is _________. [2023]

Q 19 :

The radius of fifth orbit of the ${Li}^{+ +}$ is _______ $\times 10^{- 12} m$ .

Take: radius of hydrogen atom $= 0.51 Å$ . [2023]

Q 20 :

The ratio of wavelength of spectral lines $H_{α}$ and $H_{β}$ in the Balmer series is $\frac{x}{20}$ . The value of $x$ is ________. [2023]

Q 21 :

If 917 $Å$ be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be _________ $Å$ . [2023]

Q 24 :

As per given figure A, B and C are the first, second and third excited energy level of hydrogen atom respectively. If the ratio of the two wavelengths (i.e., $\frac{λ_{1}}{λ_{2}}$ ) is $\frac{7}{4 n}$ , then the value of $n$ will be _______. [2023]

Q 25 :

The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly ______ nm. [2026]

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 1 : The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is [2024]

Q 2 : The longest wavelength associated with Paschen series is (Given RH=1.097×107 in SI unit) [2024]

Q 3 : If the wavelength of the first member of Lyman series of hydrogen is λ. The wavelength of the second member will be [2024]

Q 4 : The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly [2024]

Q 5 : The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is 915 Å.The longest wavelength of spectral lines in the Balmar series will be ________ Å. [2024]

Q 7 : A particular hydrogen like ion emits the radiation of frequency 3×1015 Hz when it makes transition from n=2 to n=1. The frequency of radiation emitted in transition from n=3 to n=1 is x9×1015 Hz, when x= __________ . [2024]

Q 9 : The number of spectral lines emitted by atomic hydrogen that is in the 4th energy level, is [2025]

Q 10 : For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is. [2025]

Q 11 : The energy levels of an atom is shown in figure. Which one of these transitions will result in the emission of a photon of wavelength 124.1 nm? Given (h=6.62×10-34 Js) [2023]

Q 12 : An electron of a hydrogen like atom, having Z = 4, jumps in from 4th energy state to 2nd energy state. The energy released in this process, will be: (Given Rch=13.6 eV) Where R = Rydberg constant c = Speed of light in vacuum h = Planck's constant [2023]

Q 13 : The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is [2023]

Q 14 : The angular momentum for the electron in Bohr’s orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be [2023]

Q 15 : The energy of He+ ion in its first excited state is. [2023] (The ground state energy for the Hydrogen atom is −13.6 eV)

Q 16 : A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be [2023]

Q 17 : For hydrogen atom, λ1 and λ2 are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of λ1 and λ2 is x32. The value of x is _________. [2023]

Q 19 : The radius of fifth orbit of the Li++ is _______ ×10-12 m. Take: radius of hydrogen atom =0.51 Å. [2023]

Q 20 : The ratio of wavelength of spectral lines Hα and Hβ in the Balmer series is x20. The value of x is ________. [2023]

Q 21 : If 917 Å be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be _________ Å. [2023]

Q 22 : A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is x×1015 Hz. The value of x is _______. (Given h=4.25×10-15 eV s) [2023]

Q 24 : As per given figure A, B and C are the first, second and third excited energy level of hydrogen atom respectively. If the ratio of the two wavelengths (i.e., λ1λ2) is 74n, then the value of n will be _______. [2023]

Q 25 : The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly ______ nm. [2026]

Want Us to Email you About Special Offers & Updates?

Q 1 :

The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is [2024]

Q 2 :

The longest wavelength associated with Paschen series is

$(Given R_{H} = 1.097 \times 10^{7} in SI unit)$ [2024]

Q 3 :

If the wavelength of the first member of Lyman series of hydrogen is $λ$ . The wavelength of the second member will be [2024]

Q 4 :

The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly [2024]

Q 5 :

The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is 915 $Å$ .The longest wavelength of spectral lines in the Balmar series will be ________ $Å$ . [2024]

Q 7 :

A particular hydrogen like ion emits the radiation of frequency $3 \times 10^{15}$ Hz when it makes transition from $n = 2$ to $n = 1$ . The frequency of radiation emitted in transition from $n = 3$ to $n = 1$ is $\frac{x}{9} \times 10^{15}$ Hz, when $x =$ __________ . [2024]

Q 9 :

The number of spectral lines emitted by atomic hydrogen that is in the $4^{th}$ energy level, is [2025]

Q 10 :

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is. [2025]

Q 11 :

The energy levels of an atom is shown in figure. Which one of these transitions will result in the emission of a photon of wavelength 124.1 nm? Given ( $h = 6.62 \times 10^{- 34} Js$ ) [2023]

Q 12 :

An electron of a hydrogen like atom, having Z = 4, jumps in from $4^{th}$ energy state to $2^{nd}$ energy state. The energy released in this process, will be: (Given $R_{c h} = 13.6 eV$ ) Where R = Rydberg constant c = Speed of light in vacuum h = Planck's constant [2023]

Q 13 :

The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is [2023]

Q 14 :

The angular momentum for the electron in Bohr’s orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be [2023]

Q 15 :

The energy of ${He}^{+}$ ion in its first excited state is. [2023]

(The ground state energy for the Hydrogen atom is −13.6 eV)

Q 16 :

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be [2023]

Q 17 :

For hydrogen atom, $λ_{1}$ and $λ_{2}$ are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of $λ_{1}$ and $λ_{2}$ is $\frac{x}{32}$ . The value of $x$ is _________. [2023]

Q 19 :

The radius of fifth orbit of the ${Li}^{+ +}$ is _______ $\times 10^{- 12} m$ .

Take: radius of hydrogen atom $= 0.51 Å$ . [2023]

Q 20 :

The ratio of wavelength of spectral lines $H_{α}$ and $H_{β}$ in the Balmer series is $\frac{x}{20}$ . The value of $x$ is ________. [2023]

Q 21 :

If 917 $Å$ be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be _________ $Å$ . [2023]

Q 24 :

As per given figure A, B and C are the first, second and third excited energy level of hydrogen atom respectively. If the ratio of the two wavelengths (i.e., $\frac{λ_{1}}{λ_{2}}$ ) is $\frac{7}{4 n}$ , then the value of $n$ will be _______. [2023]

Q 25 :

The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly ______ nm. [2026]