Frequency of the de-Broglie wave of electron in Bohr's first orbit of hydrogen atom is 1013 Hz (nearest integer). [Given : RH (Rydberg constant) = 2.18 10-18 J, h (Planck's constant) = 6.6 10-34 J. s.]
Answer
661
Solution
= v with = hmv ⇒ = mv2h = 2 KEh
∵ in the first Bohr orbit KE = RH = 2.18 10-18 J
⇒ = 2 2.18 10-186.6 10-34
= 660.6 1013 Hz 661 1013 Hz
Answer: 661
Note: The official NTA answer key gives 658. Our faculty team has re-checked this question and finds 661 correct, as explained above.