Question

The angular momentum of an electron in Bohr's orbit of He+ is 3.1625*10-34 kg-m2/sec. What is the wave number in terms of Rydberg constant (R) of the spectral line emitted an electron falls from this level to the first excited state.

(a) 3R
(B) 5R/9
(C) 3R/4
(D) 8R/9

Answer 1

The wave number in terms of Rydberg constant (R) of the spectral line emitted when an electron falls from the specified level to the first excited state is equal to 5R/9.

• The angular momentum of an electron in nth orbit of a singly occupied atom is given nh/2*pi.
• So, nh/2*pi = 3.1625*10-34 ⇒ n = 3.1625*10-34×2pi/h ⇒n=3.
• The value of n for the first excited state is equal to 2.
• Hence, for the transition, n(i) = 3 and n(f)=2.
• For helium the value of Z(atomic number ) is equal to 2.
• The wave number is given by RZ²((1/n(f)²)-(1/n(i)²)).
• substituting all the values , we get wave number to be equal to 5R/9.