Question

If the shortest wavelength of hydrogen atom in Lyman serie is x,then longest wavelength in paschen series of ${He}^{+}$ is?

• A. $\frac{9x}{5}$
• B. $\frac{36x}{5}$
• C. $\frac{36x}{7}$
• D. $\frac{5x}{9}$

Answer 1

Answer: (C)

$\frac{36x}{7}$

As we know that,

$\frac{1}{\lambda }=R(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2})Z^2$

The shortest wavelength in $H$-atom of Lyman series is from $n_1=1\to n_2=\infty$.

Given that the shortest wavelength of hydrogen atom in Lyman series is $x$.

$\therefore \frac{1}{x}=R(\frac{1}{1}-\frac{1}{\infty })1^2$

$\Rightarrow \frac{1}{x}=R$

The longest wavelength in Paschen series of ${He}^{+}$ ion will be from $n_1=3\to n_2=4$.

$\therefore \frac{1}{{\lambda }^{\prime }}=R(\frac{1}{3^2}-\frac{1}{4^2})2^2$

$\Rightarrow \frac{1}{{\lambda }^{\prime }}=\frac{4}{x}(\frac{1}{9}-\frac{1}{16})$

$\Rightarrow \frac{1}{{\lambda }^{\prime }}=\frac{4}{x}\times \left(\frac{7}{144}\right)$

$\Rightarrow {\lambda }^{\prime }=\frac{144x}{28}$

$\Rightarrow {\lambda }^{\prime }=\frac{36x}{7}$

Hence the correct answer is $\frac{36x}{7}$.