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Physics Question on Photoelectric Effect

A proton moving with a momentum P1P _{1} has a kinetic energy 18\frac{1}{8} th of its rest mass energy. Another light photon having energy equal to the kinetic energy of the possesses a momentum P2.P _{2} . Then the ratio P1−P2P1\frac{ P _{1}- P _{2}}{ P _{1}} is equal to

A

1

B

14\frac{1}{4}

C

12\frac{1}{2}

D

34\frac{3}{4}

Answer

34\frac{3}{4}

Explanation

Solution

For proton v2=c24v=c2v ^{2}=\frac{ c ^{2}}{4} v =\frac{ c }{2}
P=2mEkP =\sqrt{2 mE _{ k }}
=2̸m1̸84mc2=\sqrt{\not 2 m \frac{1}{\not 8_{4}} mc ^{2}}
P1=mc2P_{1} =\frac{ mc }{2}
For photon
Ephoton =KEproton E _{\text {photon }}= KE _{\text {proton }}
hc̸λ=18mc2̸\frac{ h \not c}{\lambda}=\frac{1}{8} mc ^{\not 2}
Pphoton =hλ=mc8=P2P _{\text {photon }}=\frac{ h }{\lambda}=\frac{ mc }{8}= P _{2}
P1−P2P1=m̸c2−m̸c8m̸c2\frac{P_{1}-P_{2}}{P_{1}}=\frac{\frac{\not m c}{2}-\frac{\not m c}{8}}{\frac{\not m c}{2}}
=12−1812=4−1812=\frac{\frac{1}{2}-\frac{1}{8}}{\frac{1}{2}}=\frac{\frac{4-1}{8}}{\frac{1}{2}}
=3×28=34=\frac{3 \times 2}{8}=\frac{3}{4}