Question

The ratio of the wavelength of photon corresponding to the $\alpha -$line of Lyman series in $H-$atom and $\beta -$line of Balmer series in $H{e}^{+}$ is:

• A. $1:1$
• B. $1:2$
• C. $1:4$
• D. $3:16$

Answer 1

The correct option is A $1:1$

$\begin{array}{c}\frac{{\lambda }_H}{{\lambda }_{He}}=\frac{R_H{z}^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)}{R_H\left(\frac{1}{1^2}-\frac{1}{2^2}\right)}\Rightarrow \frac{4\left(\frac{4-1}{16}\right)}{\left(\frac{4-1}{4}\right)}=\frac{1}{1}\\ {\lambda }_H:{\lambda }_{He}\\ 1:1\end{array}$

Hence, the correct option is $\text{A}$