An electron jumps from the 4^ th orbit to the 2^ nd orbit of hydrogen atoms. Given the Rydberg's constant R _ H =10^7 ~m ⁻¹ . frequency in Hz of the emitted radiation is ( c =3 10^8 ~m / s . )

Step-by-step solution

The correct option is (D). On plugging equation (1) in equation (2), f = cR _ H ( 1 n ₁^2 - 1 n ₂^2 ) Given, c =3 10^8 ~m / s , R _ H =10^7 ~m ⁻¹, n ₁=2 and n ₂=4 . f =3 10^8 ~ms ⁻¹ 10^7 ~m ⁻¹ ( 1 2^2 - 1 4^2 ) On solving, f = 9 16 10¹⁵ ~Hz