Question

If the shortest wavelength in Lyman series of $H$ atom is $x$, then longest wavelength in Balmer series of $H{e}^{+}$ is?

• A. $\frac{9x}{5}$
• B. $\frac{36x}{5}$
• C. $\frac{x}{4}$
• D. $\frac{5x}{9}$

Answer 1

The correct option is D $\frac{5x}{9}$

longest wavelength(${\lambda }_l$) in Balmer series means 3$\to$ 2 transition

and shortest wavelengtha(${\lambda }_s$) in Lyman series means infinte orbit $\to$ 1 transition which is equal to $x$

According to question ${\lambda }_s=R_h[\frac{1}{1^2}-\frac{1}{i^2}]=R_h=x$ ( $i=$ infinite )

${\lambda }_l=R_h[\frac{1}{2^2}-\frac{1}{3^2}]z^2=R_h\left[\frac{5}{36}\right]4=R_h\frac{5}{9}=\frac{5x}{9}$