Question

If wavelength of the photon emitted due to transition of an electron from the third orbit to the first orbit in a hydrogen atom is ${}^{\prime }\lambda {}^{\prime }$, then the wavelength of photon emitted due to transition of electron from the fourth orbit to the second orbit of a hydrogen atom will be

• A. $\frac{128}{27}\lambda$
• B. $\frac{25}{9}\lambda$
• C. $\frac{36}{7}\lambda$
• D. $\frac{125}{11}\lambda$

Answer 1

The correct option is A $\frac{128}{27}\lambda$

Wavelength emitted: is given by
$\frac{1}{\lambda }=R(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2})$
Transition from third to first gives wavelength $\frac{1}{R(\frac{1}{1^2}-\frac{1}{3^2})}=\frac{9}{8R}$ which here, is denoted by $\lambda$
Now, transition from fourth to second gives $\frac{1}{R(\frac{1}{2^2}-\frac{1}{4^2})}=\frac{16}{3R}$

Wavelength of photon due to transition of electron from fourth orbit to second orbit is $=\frac{16}{3}\times \frac{8\lambda }{9}$
$=\frac{128}{27}\lambda$