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=10−9m,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.
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Detailed Solution
1λa=R1−1161λe=R1m2−142 Given :λaλe=1m2−1421516=15⇒1m2=142+316⇒1m2=14⇒m=2 Now, we know that Kinetic energy ∝1n2KmK1=122×1=14 Now ,13.614−116=1242λe⇒13.6316=1242λe⇒λe=487nm