Question

An electron jumps from the fourth orbit to the second orbit of hydrogen atom. If Rydberg’s constant R is equal to 10⁷ m^–1, then the frequency of the emitted radiation in Hz will be

• (A) (3/16) x 10⁵
• (B) (3/6) x 10⁵
• (C) (9/16) x 10¹⁵
• (D) (3/4) x 10¹⁵

Answer 1

The Correct Option is C

Explanation:
$\frac{1}{\lambda }=R[\frac{1}{2^2}-\frac{1}{4^2}]$$\frac{v}{c}=R\left[\frac{3}{16}\right]$$\nu =3\times {10}^8\times {10}^7\times \frac{3}{16}$$=\frac{9}{16}\times {10}^{15}$