Q66 / 75 · JEE Main 2020 (09 Jan Shift 2) · Physics

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

A
$24.2 n m$
Correct answer
B
$11.4 n m$
C
$35.8 n m$
D
$8.6 n m$

Solution

\frac{h c}{λ} = (13 . 6 e V) z^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]

n_{1} = 1

n_{2} = 3

\frac{h c}{λ} = (13.6 e V) (3^{2}) [\frac{1}{1^{2}} - \frac{1}{3^{2}}]

\Rightarrow \frac{h c}{λ} = (13.6 e V) (9) (\frac{8}{9})

Wavelength

= \frac{1240}{8 \times 13.6} nm

λ = 11.39 nm

Hence, the correct answer is option: 'B'.

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