CBSE Class 12 · 2019 · Modern PhysicsShort Answer

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Obtain the expression for the ratio of the de-Broglie wavelengths associated with the electron orbiting in the second and third excited states of a hydrogen atom.

Answer: $λ_{3} : λ_{4} = 3 : 4$ (second & third excited states).

Solution

For a Bohr orbit,

2 π r_{n} = n λ_{n}

, and

r_{n} \propto n^{2}

, so

λ_{n} = \frac{2 π r _{n}}{n} \propto n .

Second excited state is

n = 3

and third excited state is

n = 4

\frac{λ _{(n = 3)}}{λ _{(n = 4)}} = \frac{3}{4} .

λ_{3} : λ_{4} = 3 : 4

CBSE Class 12 Board · 2019 · Subjective

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