Question

(viii) In hydrogen atom, electron jumps from the 3rd orbit to the
1st orbit. The change in angular momentum is
(a) 1.05x10-34 Js (b) 2.11x10-34 Js
(c) 3.16*10-34 Js (d) 4.22x10-34 Js

Answer 1

Answer:

we know that the general formula for possible angular momentum is L=$\frac{nh}{2pi }$ where n is principal quantum number and h is planck constant, hence by inserting our  value, the answer will be B

Explanation:

Answer 2

In hydrogen atom, when an electron jumps from the third orbit to the first orbit, The change in angular momentum is equal to 2.11x10-34 Js.

• We know from Bohr's quantization theory of angular momentum that the angular momentum of an electron in nth orbit is equal to nh/2π.
• So, angular momentum of electron in third orbit is equal to 3h/2π and the angular momentum of electron in first orbit is equal to 1h/2π.
• Therefore, the difference in angular momentum is equal to (3-1)h/2π = h/π .
• Taking the value of h to be 6.626×10-34 Js and that of π to be 3.14 , we get the difference in angular momentum to be 2.11x10-34 Js.