JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Modern Physics

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number

n_{i}

to another with quantum number

n_{f} . V_{i}

and

V_{f}

are respectively the initial and final potential energies of the electron. If

\frac{V _{i}}{V _{f}} = 6.25

, then the smallest possible

n_{f}

Answer: 5

Solution

In a hydrogen atom the potential energy is

V_{n} = - \frac{k e ^{2}}{r _{n}}

with

r_{n} \propto n^{2}

, so

V_{n} \propto - \frac{1}{n ^{2}}

. Then

\frac{V _{i}}{V _{f}} = \frac{n _{f}^{2}}{n _{i}^{2}} = 6.25 = \frac{25}{4}

, giving

\frac{n _{f}}{n _{i}} = \frac{5}{2}

. The smallest integers in this ratio are

n_{i} = 2

and

n_{f} = 5

, so the smallest possible

n_{f}

5

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Integer answer

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