JEE Advanced 2013 · Paper 2 · Physics Q4

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PhysicsMultiple correct+3 / -1Question 4 of 60

The radius of the orbit of an electron in a Hydrogen-like atom is $4.5a_0$ , where $a_0$ is the Bohr radius. Its orbital angular momentum is $\dfrac{3h}{2\pi}$ . It is given that $h$ is Planck constant and $R$ is Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are)

\dfrac9{32R}

\dfrac9{16R}

\dfrac9{5R}

\dfrac4{3R}

Answer and worked solution

Official answer: A, C

Quantum number. $L=\dfrac{nh}{2\pi}=\dfrac{3h}{2\pi}$ , so $n=3$ .

Z. $r=\dfrac{n^2a_0}Z=4.5a_0$ gives $Z=2$ .

Transitions ( $\frac1\lambda=Z^2R\left(\frac1{n_1^2}-\frac1{n_2^2}\right)$ ):

3 → 1: $4R\cdot\frac89=\frac{32R}9$ , $\lambda=\frac9{32R}$ ✓
3 → 2: $4R\cdot\frac5{36}=\frac{5R}9$ , $\lambda=\frac9{5R}$ ✓
2 → 1: $3R$ , $\lambda=\frac1{3R}$ (not listed)

(A), (C)

Topics:Bohr model and hydrogen spectrum

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JEE Advanced 2013 · Paper 2, Physics question 4