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Q.

Radiation given out during transition of n = 2 to n = 1 for H-atoms are allowed to fall on He + ions in n = 1 and n = 2 states. The possible transitions of helium ion due to absorption of energy is :

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a

n=2 to n=4

b

n=2 to n=3

c

n=1 to n=4

d

n=2 to n=5

answer is D.

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Detailed Solution

$∆$ E given out during H-atom deexcitation from n = 2 to n = 1
$\begin{array}{l} ΔE = \frac{ch}{λ} = ch \cdot R_{H} [\frac{1}{1^{2}} - \frac{1}{2^{2}}] \\ = ch \times R_{H} \times \frac{3}{4} \end{array}$
For needed He atom excitation from n = 2 to n = 4
$\begin{aligned} ΔE = \frac{hc}{λ} = ch \cdot R_{H} [\frac{1}{2^{2}} - \frac{1}{4^{2}}] \times Z^{2} \\ = ch \times R_{H} \times \frac{3}{16} \times 2^{2} = ch \cdot R_{H} \times \frac{3}{4} \end{aligned}$

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