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

Answer 1

Answer: (C)

(C) (9/16) x 1015

Answer 2

Explanation:
1λ=R[122−142]vc=R[316]ν=3×108×107×316=916×1015