Question

If the shortest wavelength in Lyman series of hydrogen atom is $A$, then the longest wavelength in Paschen series of $H{e}^{+}$ is :

• A. $\frac{5A}{9}$
• B. $\frac{9A}{5}$
• C. $\frac{36A}{5}$
• D. $\frac{36A}{7}$

Answer 1

Answer: (D)

Shortest wavelength is corresponding to best ine
n_L = 1 (Lyman series)
n_H = $\infty$ (infinite)
$\frac{1}{A}=r\times {\left(1\right)}^2\left\{\frac{1}{12}-\frac{1}{2}\right\}=R$
Longest wavelength = 1^st Line
= 3 n_H = 4
$\frac{1}{\lambda }=r\times {\left(2\right)}^2\left\{\frac{1}{3^2}-\frac{1}{4^2}\right\}=\frac{r\times 7}{36}$
$\lambda =\frac{36A}{7}$