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A photon with initial frequency 10^(11) ...

A photon with initial frequency `10^(11)` Hz scatters off electron at rest. Its final frequency is `0.9 xx 10^(11)` Hz. The speed of scattered electron is close to: `(h = 6.63 xx 10^(-34) Js. M_(e) = 9.1xx10^(-31) kg)`

A

`4xx10^(3)ms^(-1)`

B

`3xx10^(2)ms^(-1)`

C

`2xx10^(6)ms^(-1)`

D

`30 ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
Initial energy of photon = Energy of photon after scattering + K.E. of scattered electron
`hv_(1)=hv_(2)+(1)/(2) mv^(2) rArr (1)/(2) mv^(2) =h(v_(1)-v_(2))`
`v^(2)=(2h(v_(1)-v_(2)))/(m)=(2xx6.63xx10^(-34)(10^(11)-0.9xx10^(11)))/(9.1xx10^(-31))`
`v= sqrt(14615384)=3.8xx10^(3)~~4xx10^(3)ms^(-1)`
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