Question

If the shortest wavelength of hydrogen atom in Lyman series is $x$, then longest wavelength in Paschen series of $H{e}^{+}$ is

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

Answer 1

Answer: (C)

For shortest wavelength line (maximum energy), $n_1=1,n_2=\infty ,Z=1$

$\frac{1}{{\lambda }_{\min }}=R\cdot Z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

$\Rightarrow \frac{1}{x}=R.1^2\left[\frac{1}{1^2}-\frac{1}{{\infty }^2}\right]=R$

For Paschen series line having longest wavelength (minimum energy $),n_1=3,n_2=4$

For $H{e}^{+}$

$\frac{1}{{\lambda }_{\max }}=R\times 4\left[\frac{1}{3^2}-\frac{1}{4^2}\right]$

$\frac{1}{{\lambda }_{\max }}=\frac{1}{x}\cdot 4\left[\frac{1}{9}-\frac{1}{16}\right]$

$=\frac{1}{x}\times 4\times \left[\frac{16-9}{9\times 16}\right]$

$=\frac{7\times 4}{16\times 9\times x}$

$\frac{1}{{\lambda }_{\max }}=\frac{7}{36x}$

$\Rightarrow {\lambda }_{\max .}=\frac{36x}{7}$