Question

What is the wavelength transition corresponding to last line of brackett series in $H{e}^{+}$ ion?

• A. $\frac{25}{8R}$
• B. $\frac{4}{R}$
• C. $\frac{1}{R}$
• D. $\frac{16}{R}$

Answer 1

Answer: (B)

$\frac{4}{R}$

$n_1=4$ $n_2=\infty$

last line of Brackett series in $H{e}^{+}$ ion

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

$\frac{1}{\lambda }=R(\frac{1}{4^2}-\frac{1}{6})2^2$

$\frac{1}{\lambda }=\frac{R}{4}$

$\lambda =\frac{4}{R}$.