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Q.

The energy required to ionize a hydrogen like ion in its ground state is 9 Rydberg. What is the wavelength of the radiation emitted when the electron In this ion jumps from the second excited state to the ground state?

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a

11.4nm

b

35.8nm

c

8.6nm

d

24.2nm

answer is D.

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Detailed Solution

Rydberg energy=13.6 eV.

So, ionization energy= $(13.6 z^{2}) e V = 9 \times 13.6 e V$
$\Rightarrow Z = 3 W h e n e l e c t r o n j u m p s f r o m o r b i t 3 t o 1, \frac{1}{λ} = R Z^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
$\Rightarrow \frac{1}{λ} = 1.09 \times 10^{7} \times 9 [\frac{1}{1} - \frac{1}{3^{2}}] \Rightarrow λ = 11.4 n m$

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