JEE Advanced 2017 · Paper 1 · Physics Q9

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PhysicsInteger+3Question 9 of 54

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 $n_f$ is

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Answer and worked solution

Official answer: 5

In the Bohr model $V_n\propto-\dfrac1{n^2}$ , so
$\frac{V_i}{V_f}=\frac{n_f^2}{n_i^2}=6.25\implies\frac{n_f}{n_i}=\frac52.$

The smallest integers that fit are $n_i=2$ and $n_f=5$ .

Answer: 5

Topics:Bohr model and hydrogen spectrum

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JEE Advanced 2017 · Paper 1, Physics question 9