Q.

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=15 , which of the option (s) is /are correct ?
[Use hc=1242eVnm;1nm=109m,h, and c are Planck’s constant and speed of light, respectively]
 

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a

Δpa/Δpe=12

b

The ratio of kinetic energy of the electron in the state n = m to the state n = 1 is 14

c

m = 2

d

 λe= 418 nm

answer is B, C.

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Detailed Solution

1λa=R11161λe=R1m2142 Given :λaλe=1m21421516=151m2=142+3161m2=14m=2 Now, we know that Kinetic energy 1n2KmK1=122×1=14 Now ,13.614116=1242λe13.6316=1242λeλe=487nm

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