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]

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

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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]

0%

(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]

0%

(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]

0%

(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 aom, the wavelength of emitted radiation is 2000 $\overset{°}{A}$ and it becomes 6000 $\overset{°}{A}$ when the electron jumps from state B to stacte 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)

Figure

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 wavbelength 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)

Figure

Largest wavelength of Balmer series

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

$λ_{2} = \frac{36}{5 R}$

Figure

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

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]

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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]

0%

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]

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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]

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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]

0%

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]

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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]