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JEE Advanced 2019 Paper 2

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Question : 1

Total: 54

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

, 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]

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

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