JEE PHYSICS MCQ: ATOMS

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Analyzing +4 | -1

An electron in hydrogen atom first jumps from second excited state to first excited state and then from first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons emitted in these two cases be a, b and c respectively. Then
(i)

\mathrm{c}=1 / \mathrm{a}\quad

(ii)

\mathrm{a}=9 / 4

(iii)

\mathrm{b}=5 / 27\quad

(iv)

\mathrm{c}=5 / 27

Correct options are -

View answer and explanation

Correct option: A

i, iii, iv

Option A is correct because c = 1/a, b = 5/27, and c = 5/27 are the correct relationships between wavelength, momentum, and energy of the photons emitted.

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