JEE Advanced 2019 Paper 2

Section: Physics

Question : 1 of 54

Marks: +1, -0

A free hydrogen atom after absorbing a photon of wavelength λa gets excited from the state n = 1 to the state n = 4. Immediately after that the electron jumps to n=m state by emitting a photon of wavelength λe.Let the change in momentum of atom due to the absorption and the emission are Δpa and Δpe , respectively. If

λa

λe

, which of the option (s) is /are correct ?
[Use hc=1242‌ev‌nm;1‌nm‌10−9] m, h and c are Planck’s constant and speed of light, respectively]

[JEE Adv 2019 P2]

Δpa/Δpe=
1
2
The ratio of kinetic energy of the electron in the state n = m to the state n = 1 is
1
4
m=2
λe=418nm

Validate

Go to Question:

Next Question