Question

The longest wavelength of $H{e}^{+}$ in Paschen series is "m", then shortest wavelength of $B{e}^{+3}$ in Paschen series is (in terms of m):

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

Answer 1

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

The expression for the wavelength is $\frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

For the longest wavelength of the He+ atom in Paschen series , the expression becomes $\frac{1}{m}=R{2}^2[\frac{1}{3^2}-\frac{1}{4^2}]$......(1)

For the shortest wavelength of $B{e}^{+3}$, the expression becomes

$\frac{1}{\lambda }=R{4}^2[\frac{1}{3^2}-\frac{1}{{inf}^2}]$......(2)
Divide equation (2) with equation (1).

$\frac{\frac{1}{\lambda }}{\frac{1}{m}}=\frac{R{4}^2[\frac{1}{3^2}-\frac{1}{{inf}^2}]}{R{2}^2[\frac{1}{3^2}-\frac{1}{4^2}]}$

On solving this, we get $\lambda =\frac{7}{64}m$