Question

If the wavelength of the photon emitted due to transition of an electron from the third orbit to the first orbit in a hydrogen atom is $\lambda$, then the wavelength of photon emitted due to transition of an electron from the fourth orbit to the second orbit 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 $\frac{128}{27}\lambda$

The wavelength of the emitted photon is given by -
$\frac{1}{\lambda }=R{z}^2(\frac{1}{n_2^2}-\frac{1}{n_1^2})$
Emission of 3rd orbit to 1st orbit
$\frac{1}{\lambda }=R{z}^2(\frac{1}{1^2}-\frac{1}{3^2})$
$\frac{1}{\lambda }=R{z}^2(1-\frac{1}{9})=\frac{8R{z}^2}{9}...\left(i\right)$
Similarly for the emission of electron from 4th orbit to 2nd orbit
$\frac{1}{{\lambda }^{\prime }}=R{z}^2(\frac{1}{2^2}-\frac{1}{4^2})$
$\frac{1}{{\lambda }^{\prime }}=R{z}^2(\frac{1}{4}-\frac{1}{16})$
$\frac{1}{{\lambda }^{\prime }}=\frac{3R{z}^2}{16}...\left(ii\right)$
Divide (i) by (ii)
$\frac{1}{\lambda }\times \frac{{\lambda }^{\prime }}{1}=\frac{8R{z}^2}{9}\times \frac{16}{3R{z}^2}$
${\lambda }^{\prime }=\frac{128\lambda }{27}$