JEE Main Atom PYQ

1. (JEE Main 2023 (Online) 12th April Morning Shift)

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

A. 2

B. 4

C. 3

D. 1

The Correct Answer is Option (C)

The energy of an electron in an excited state of hydrogen is given by the equation:

Code snippet

$E = - 13.6 \frac{1}{n^{2}} eV$

where n is the principal quantum number of the state.

The energy of the electron beam is 12.5 eV, which is enough to excite the electron to the n = 3 state.

The possible transitions from n = 3 to lower energy states are:

n = 3 to n = 2, with a wavelength of 656.33 nm (H-alpha)

n = 3 to n = 1, with a wavelength of 102.57 nm (Lyman-alpha)

n = 2 to n = 1, with a wavelength of 121.57 nm (Lyman-beta)

Therefore, there are 3 possible spectral lines that can be emitted.

2. (JEE Main 2023 (Online) 6th April Morning Shift)

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

JEE Main 2023 (Online) 6th April Morning Shift Physics - Atoms and Nuclei Question 4 English

A. A

B. C

C. B

D. D

The Correct Answer is Option ()

Currently no explanation available

3. (JEE Main 2023 (Online) 1st February Evening Shift)

An electron of a hydrogen like atom, having $Z = 4$ , jumps from $4^{th}$ energy state to $2^{nd}$ 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

A. $10.5 eV$

B. $40.8 eV$

C. $13.6 eV$

D. $3.4 eV$

The Correct Answer is Option (B)

The energy difference between the 4th and 2nd energy states of a hydrogen-like atom can be calculated using the formula for the energy levels of a hydrogen-like atom:

$\begin{aligned} Δ E & = 13.6 Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) \\ Z & = 4 (hydrogen like atom) \\ n_{1} & = 2, n_{2} = 4 \\ Δ E & = 13.6 (4)^{2} (\frac{1}{4} - \frac{1}{16}) \\ = 13.6 \times (\frac{16 - 4}{64}) \times 16 \\ Δ E & = 13.6 \times \frac{12}{64} \times 16 \\ Δ E & = 40.8 eV \end{aligned}$

4. (JEE Main 2023 (Online) 25th January Evening Shift)

The energy levels of an atom is shown in figure.

JEE Main 2023 (Online) 25th January Evening Shift Physics - Atoms and Nuclei Question 31 English

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)

A. C

B. B

C. A

D. D

The Correct Answer is Option (D)

$\begin{aligned} λ = \frac{hc}{Δ E} \\ Δ E_{A} = 2.2 eV \\ Δ E_{B} = 5.2 eV \\ Δ E_{C} = 3 eV \\ Δ E_{D} = 10 eV \\ λ_{A} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{2.2 \times 1.6 \times 10^{- 19}} \\ = \frac{12.41 \times 10^{- 7}}{2.2} m \\ = \frac{1241}{2.2} nm = 564 nm \\ λ_{B} = \frac{1241}{5.2} nm = 238.65 nm \\ λ_{C} = \frac{1241}{3} nm = 413.66 nm \\ λ_{D} = \frac{1241}{10} = 124.1 nm \end{aligned}$

5.(JEE Main 2023 (Online) 24th January Evening Shift)

A photon is emitted in transition from n = 4 to n = 1 level in hydrogen atom. The corresponding wavelength for this transition is (given, h = 4 $\times$ 10 $^{- 15}$ eVs) :

A. 99.3 nm

B. 94.1 nm

C. 974 nm

D. 941 nm

The Correct Answer is Option (B)

$\frac{h c}{λ} = + 13.6 eV [\frac{1}{1} - \frac{1}{4^{2}}]$

$\Rightarrow \frac{4 \times 10^{- 15} \times 3 \times 10^{- 8}}{λ} = 13.6 [\frac{15}{16}]$

$λ = 94.1 nm$