Question

The longest wavelength of ${He}^{+}$ in Paschen series is '$m$' then shortest wavelength of ${Be}^{3+}$ in Paschen series in terms of '$m$' is:

• A. $\frac{7}{64}m$
• B. $\frac{5}{36}m$
• C. $\frac{64}{7}m$
• D. $\frac{53}{8}m$

Answer 1

The correct option is A $\frac{7}{64}m$

$\frac{1}{{\lambda }_{{He}^{+}}}=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\frac{1}{m}=R\times 4[\frac{1}{9}-\frac{1}{16}]=\frac{7R}{36}$ ...(i)

$\frac{1}{{\lambda }_{{Be}^{3+}}}=R\times 16[\frac{1}{9}-\frac{1}{\infty }]=\frac{16R}{9}$ ....(ii)

Dividing equation (i) by (ii)

$\frac{{\lambda }_{{Be}^{3+}}}{m}=\frac{7\times 9}{16\times 36}$

${\lambda }_{{Be}^{3+}}=\frac{7}{64}m$.

Hence, the correct option is $\text{A}$