An electron in a hydrogen atom is excited from the ground state to a higher energy state such that its de Broglie wavelength increases by a factor of 4 . The increase in the angular momentum of the electron is (Given Planck's constant h = 6.6 10⁻³⁴ J s ):

Step-by-step solution

The de Broglie wavelength of an electron in the n^ th orbit is given by = h mv . From Bohr's quantization rule, mvr = nh 2 mv = nh 2 r . Thus, = 2 r n . Since the radius of the n^ th orbit r n^2 , we get: n^2 n n Given that the de Broglie wavelength increases by a factor of 4 , the principal quantum number n must also increase by a factor of 4 . Since the electron is initially in the ground state ( n₁ = 1 ), the final state is n₂ = 4 . The increase in angular momentum is: L = (n₂ - n₁) h 2 = (4 - 1) h 2 = 3 6.6 10⁻