Question

An electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom. Given the Rydberg's constant R = 10⁵ cm⁻¹. The frequency in Hz of the emitted radiation will be
(a) $\frac{3}{16} \times10^5$
(b) $\frac{3}{16} \times10^{15}$
(c) $\frac{9}{16} \times10^{15}$
(d) $\frac{3}{4} \times10^{15}$

Answer 1

option A is correct.......

Answer 2

Answer:

c) 9/ 16 × 10`5

Explanation:

Orbits = 2 (Given)

Rydberg's constant R = 10⁵ cm⁻¹ (Given)

Transfer - fourth orbit to second orbit of hydrogen atom (Given)

1/λ = R ( 1/2² - 1/4²) = 3R/16

= λ = 3R/16

= 16/3 × 10`5

Frequency = n = c/λ

= 3 × 10`10/ 16/3 × 10`5

= 9/ 16 × 10`5

Thus, the frequency in Hz of the emitted radiation will be 9/ 16 × 10`5.