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Q.

An electron of a hydrogen like atom, having Z = 4, jumps 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

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a

13.6 eV

b

10.6 eV

c

3.4 eV

d

40.8 eV

answer is D.

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Detailed Solution

ΔE=13.6Z2122−142

=13.642316=13.6×3

= 40.8 eV

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