JEE Main Atom PYQ

11. (JEE Main 2023 (Online) 25th January Morning Shift)

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 $\frac{20}{x} λ_{0}$ . The value of $x$ is _____________.

The Crrect Answer is

Second excited state $\to$ first excited state , $n = 3$ to $n = 2$

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

Third excited state $\to$ second orbit , $n = 4$ to $n = 2$

$\frac{1}{λ} = R (\frac{1}{4} - \frac{1}{16}) = (\frac{3}{16} R)$

Taking ratio of (1) and (2)

$\frac{λ}{λ_{0}} = \frac{5}{36} \times \frac{16}{3} = (\frac{20}{27})$

$λ = \frac{20}{27} λ_{0}$

$x = 27$

12. ( JEE Main 2022 (Online) 29th July Morning Shift)

Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.

A. 3 : 4

B. 4 : 3

C. 1 : 4

D. 4 : 1

The Correct Answer is Option (A)

$E_{1} = E_{0} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = E_{0} \times \frac{3}{4}$

$E_{2} = E_{0}$

$∴$ $\frac{E_{1}}{E_{2}} = \frac{3}{4}$

13. (JEE Main 2022 (Online) 25th July Evening Shift)

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength $λ$ . The value of principal quantum number ' $n$ ' of the excited state will be : ( $R :$ Rydberg constant)

A. $\sqrt{\frac{λ R}{λ - 1}}$

B. $\sqrt{\frac{λ R}{λ R - 1}}$

C. $\sqrt{\frac{λ}{λ R - 1}}$

D. $\sqrt{\frac{λ R^{2}}{λ R - 1}}$

The Correct Answer is Option (B)

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

$\Rightarrow \frac{1}{λ R} = 1 - \frac{1}{n^{2}}$

$\Rightarrow \frac{1}{n^{2}} = 1 - \frac{1}{λ R} = \frac{λ R - 1}{λ R}$

$\Rightarrow n = \sqrt{\frac{λ R}{λ R - 1}}$

14. (JEE Main 2022 (Online) 30th June Morning Shift)

A hydrogen atom in ground state absorbs 12.09 eV of energy. The orbital angular momentum of the electron is increased by :

A. 1.05 $\times$ 10^{$-$ 34} Js

B. 2.11 $\times$ 10^{$-$ 34} Js

C. 3.16 $\times$ 10^{$-$ 34} Js

D. 4.22 $\times$ 10^{$-$ 34} Js

The Correct Answer is Option (B)

Change in energy

$\begin{aligned} Δ E = E_{f} - E_{2} \Rightarrow 12.09 = E_{f} - (- 13.6) \Rightarrow E_{f} = - 1.51 eV \\ \Rightarrow \frac{- 13.6}{n^{2}} = - 1.51 \Rightarrow n^{2} = 9 \Rightarrow n = 3 \\ So, Δ L = L_{f} - L_{i} \\ = \frac{h}{2 π} (3 - 1) = \frac{2 h}{2 π} = \frac{h}{π} = \frac{6.63 \times 10^{- 34}}{3.14} = 2.11 \times 10^{- 34} Js \end{aligned}$

15. (JEE Main 2022 (Online) 27th June Morning Shift)

A hydrogen atom in its ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of :

(Given, Planck's constant = 6.6 $\times$ $-$ 34 Js).

A. 2.10 $\times$ 10^{$-$ 34} Js

B. 1.05 $\times$ 10^{$-$ 34} Js

C. 3.15 $\times$ $-$ 34 Js

D. $\times$ $-$ 34 Js

The Correct Answer is Option (B)

$- 13.6 + 10.2 = \frac{- 13.6}{n^{2}}$

$\Rightarrow \frac{13.6}{n^{2}} = 3.4$

$\Rightarrow n = 2$

$\Rightarrow Δ L = 2 \times \frac{h}{2 λ} - 1 \times \frac{h}{2 λ}$

$= \frac{h}{2 λ}$

$\Rightarrow Δ L ≃ 1.05 \times 10^{- 34}$ Js