Question

The energy required to iononise 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 state, to the ground state?

• A. $24.2\text{nm}$
• B. $\text{11.4 nm}$
• C. $\text{35.8 nm}$
• D. $\text{8.6 nm}$

Answer 1

The correct option is B $\text{11.4 nm}$

According to Bohr's theory, ionization energy $(I.E)$ is

$I.E=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\Rightarrow 9R=R{Z}^2[\frac{1}{1^2}-\frac{1}{{\infty }^2}]$

$\therefore Z=3$

Now, when electron jumps from its second excited state to ground state, then

$n_1=1,n_2=3$

$\Rightarrow \frac{1}{\lambda }=R\times 3^2[\frac{1}{1^2}-\frac{1}{3^2}]$

$\therefore \frac{1}{\lambda }=9R(1-\frac{1}{9})$

$\Rightarrow \lambda =\frac{1}{8R}=\frac{1}{8\times 1.097\times {10}^7}(\because R=10973731.6{\text{m}}^{-1})$

$\Rightarrow \lambda =11.39\text{nm}$

Hence, $\left(B\right)$ is the correct answer.