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Question

Physics Question on Nuclei

A radioactive nucleus of mass MM emits a photon of frequency υ\upsilon and the nucleus recoils. The recoil energy will be

A

Mc2hυMc^2-h\upsilon

B

h2υ2/2Mc2h^2\upsilon^2 /2 Mc^2

C

zero

D

hυ\upsilon

Answer

h2υ2/2Mc2h^2\upsilon^2 /2 Mc^2

Explanation

Solution

Fex=dPdt=0F _{ ex }=\frac{ d P }{ dt }=0

dP=0\Rightarrow dP =0

P=\Rightarrow P = constant

Pi=Pf\vec{ P }_{ i }=\vec{ P }_{ f }

0=PNu+PPh0=\vec{ P }_{ Nu }+\vec{ P }_{ Ph }

PNu=PPh=hλ=hvc\left|\vec{ P }_{ Nu }\right|=\left|\vec{ P }_{ Ph }\right|=\frac{ h }{\lambda}=\frac{ hv }{ c }

Recoil K.EK.E. of nucleus K.ENu=PNu22MNuK . E _{ Nu }=\frac{ P _{ Nu }^{2}}{2 M _{ Nu }}

K.ENu=(hv/c)22M=h2v22Mc2K.E _{\cdot Nu }=\frac{( hv / c )^{2}}{2 M }=\frac{ h ^{2} v^{2}}{2 Mc ^{2}}