Question

The shortest wavelength of the Lyman series of hydrogen is equal to the shortest wavelength of the Balmer series of a hydrogen-like atom of atomic number ${}^{\prime }{Z}^{\prime }$. The value of $Z$ is

• A. $4$
• B. $2$
• C. $3$
• D. $6$

Answer 1

The correct option is B $2$

For Hydrogen, $Z=1$
Shortest wavelength of Lymen series:
$\frac{1}{\lambda }=R\left(1\right)=R$
$\lambda =\frac{1}{R}$

For hydrogen like atom with $Z$,

Balmer series shortest wavelength:
$\frac{1}{\lambda }=Z^{2}R\left(\frac{1}{2^2}\right)=\frac{Z^{2}R}{4}$

Given both the wavelengths are equal:
$Z^2=4$
$Z=2$ (negative value not possible)