Question

The wavelength of the first line of Iyman series is $\lambda$. The wavelength of first line in Balmer series is __________.

• A. $\frac{27}{5}\lambda$
• B. $\frac{5}{27}\lambda$
• C. $\frac{9}{2}\lambda$
• D. $\frac{2}{5}\lambda$

Answer 1

The correct option is A $\frac{27}{5}\lambda$

Wavelength of a electron in hydrogen atom can be claculated by $\frac{1}{\lambda }=R_H(\frac{1}{n_i^2}-$ $\frac{1}{n_f^2})$ $R_H=$Rydberg Constant

For First line of Iyman series $n_i=1,n_f=2$

$\Rightarrow \frac{1}{{\lambda }_L}=R_H(\frac{1}{1^2}-\frac{1}{2^2})=\frac{3R_H}{4}$

$\Rightarrow R_H=\frac{4}{3{\lambda }_L}$

Now, For First line of Balmer series $n_i=2,n_f=2$

$\Rightarrow \frac{1}{{\lambda }_B}=R_H(\frac{1}{2^2}-\frac{1}{3^2})=\frac{5R_H}{36}=\frac{5}{27{\lambda }_L}$

$\Rightarrow {\lambda }_B=\frac{27{\lambda }_L}{5}$

Therefore, A is correct option.