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CGP EDU Academic Team
Published on: September 12, 2026
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
Text Solution
Verified by ExpertsThe correct answer is:
D
=13.6[4 - 1] eV
= 13.6 x 3 = 40.8eV
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