JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Modern Physics

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)13.6 eV
(B)10.5 eV
(C)3.4 eV
(D)40.8 eV✓ Correct

Answer: (D)

Solution

For a hydrogen-like atom the energy released in a transition is

Δ E = 13.6 Z^{2} (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}})

eV. With

Z = 4

n_{i} = 4

and

n_{f} = 2

Δ E = 13.6 \times 16 \times (\frac{1}{4} - \frac{1}{16}) = 13.6 \times 16 \times \frac{3}{16} = 13.6 \times 3

. The energy released is

40.8

eV.

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Single correct option

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