A line in Lyman series of hydrogen atom has a wavelength of 1 . 03 × 10 - 7 m . What is the initial energy level of the electron?

For Lyman series, the transition of electron takes place from a higher energy level to ground state (i.e. n = 1 ). Hence, 1 λ = 1 . 09677 × 10 7 1 1 2 - 1 n 2 m - 1 On putting the given value of wavelength ( λ ) = 1 . 03 × 10 - 7 m in the above equation, we get, 1 1 . 03 × 10 - 7 m = 1 . 09677 × 10 7 1 1 2 - 1 n 2 m - 1 ⇒ 1 - 1 n 2 = 1 1 . 03 × 10 - 7 m × 1 . 09677 × 10 7 m - 1 ⇒ 1 n 2 = 1 - 1 1 . 13 ⇒ n = 113 13 ⇒ n ≈ 3 Therefore, for the given transition, the electron was initially at n = 3 .

If the position of the electron is measured within an accuracy of ± 0 . 002 nm . Calculate the uncertainty in the momentum of the electron.

Given: Accuracy in measurement of position, ∆ x = ± 0 .002 nm = ± 0 .002 × 10 - 9 m According to Heisenberg’s uncertainty principle, ∆ x . ∆ p ≥ h 4 π Where, ∆ x = Uncertainty in position of the electron ∆ p = Uncertainty in momentum of the electron By applying the uncertainty principle, Δp = h 4 πΔx = 6 . 626 × 10 - 34 4 × 3 . 14 × 0 . 002 × 10 - 9 = 2 . 63 × 10 - 23 Kg m / s

If the velocity of the electron is 1 . 6 × 10 6 ms - 1 , calculate de Broglie wavelength associated with this electron.

Given: Velocity of electron = 1 . 6 × 10 6 ms - 1 The mass of electron = 9 . 1 × 10 - 31 kg According to de-Broglie’s expression, λ = h mν = 6 .626 × 10 - 34 Js 9 .1 × 10 - 31 kg 1 . 6 × 10 6 ms - 1 = 455 pm Therefore, the wavelength associated with the electron is 455 pm .