Question

An electron jumps from the fourth orbit to the second orbit of hydrogen atom. Given: the Rydberg's constant $R={10}^{5}c{m}^{-1}$. The frequency, in $Hz$, of the emitted radiation will be

• A. $\frac{3}{16}\times {10}^5$
• B. $\frac{3}{6}\times {10}^{15}$
• C. $\frac{9}{16}\times {10}^5$
• D. $\frac{9}{16}\times {10}^{15}$

Answer 1

The correct option is D $\frac{9}{16}\times {10}^{15}$

Frequency is $c\overline{\nu }$.
Therefore, $\nu =3\times {10}^{10}\times {10}^5(\frac{1}{4}-\frac{1}{16})=\frac{9}{16}\times {10}^{15}{s}^{-1}$