Question

The shortest wavelength of hydrogen atom in Lyman series is $\lambda$. The longest wavelength in Balmer series of $He ^{+}$is

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

Answer 1

Answer: (A)

For $H : \frac{1}{\lambda}= R _{ H } \times 1^2\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right) .....$(1)
$\frac{1}{\lambda_{ He ^{+}}}= R _{ H } \times 2^2 \times\left(\frac{1}{4}-\frac{1}{9}\right).....$(2)
From (1) & (2) $\frac{\lambda_{ He ^{+}}}{\lambda}=\frac{9}{5}$
$ \lambda_{ He ^{+}}=\lambda \times \frac{9}{5}$
$ \lambda_{ He ^{+}}-\frac{9 \lambda}{5}$