Question

If an electron in hydrogen atom jumps from an orbit of level $n=3$ to an orbit of level $n=2$, the emitted radiation has a frequency $(R=$ Rydberg constant, $C=$ velocity of light)

• A. $\frac{3RC}{27}$
• B. $\frac{RC}{25}$
• C. $\frac{8RC}{9}$
• D. $\frac{5RC}{36}$

Answer 1

Answer: (D)

$\frac{5RC}{36}$

Wavelength of photon emitted due to electronic transition from $n_1=3$ to $n_2=2$ of hydrogen atom

$\frac{1}{\lambda }=R(\frac{1}{n_2^2}-\frac{1}{n_1^2})$

$\therefore$ $\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{3^2})$ $⟹\lambda =\frac{36}{5R}$

Thus frequency of photon $\nu =\frac{C}{\lambda }=\frac{5RC}{36}$