Question

If the shortest wavelength of the spectral line of ${He}^{+}$ in Lyman series is $x$, then longest wavelength of the line in Balmer series of ${Li}^{2+}$ is:

• A. $\frac{5x}{4}$
• B. $\frac{4x}{5}$
• C. $\frac{16x}{5}$
• D. $9x$

Answer 1

Answer: (C)

$\frac{16x}{5}$

Shortest wavelength $=n_{\infty }\to n_1$

$\frac{1}{{\lambda }_{min}}=R_H\frac{Z^2}{n_1^2}$

for ${He}^{+},Z=2$

$\frac{1}{{\lambda }_{min}}=R_H\frac{2^2}{1^2}$ or $R_H=\frac{1}{4x}$

for the longest wavelength in Balmer series of ${Li}^{2+},Z=3,n_1=2,n_2=3$

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

$\frac{1}{{\lambda }_{max}}=\frac{1}{4x}{\left(3\right)}^2(\frac{1}{4}-\frac{1}{9})$

$\frac{1}{{\lambda }_{max}}=\frac{9}{4x}\times \frac{5}{36}$

${\lambda }_{max}=\frac{16x}{5}$