A 662 kev photon from 137Cs is Compton scattered at an angle of 60 calculate the energy of the scattered photon.

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llltkl | College Teacher | (Level 3) Valedictorian

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The Compton scattering equation is:

`lambda' - lambda = (h/(m_ec))(1-costheta) `

Converting energy in ev to J, then plugging in the values, we get


`rArr lambda=(6.626*10^-34*3*10^8)/(662*10^3*1.6022*10^-19)`

`=1.874*10^-12 m`

Put this value in Compton scatter equation to get

`lambda'= lambda + (h/(m_ec))(1-costheta)`

`=1.874*10^-12+((6.626*10^-34) (1-1/2))/(9.1094*10^-31*3*10^8)`

`=(1.874+1.2123)*10^-12 m`

`=3.08642*10^-12 m`

Energy of the scattered photon

`=(6.626*10^-34*3*10^8)/( 3.08642*10^-12 *1.6022*10^-19)`

`=401977 ev`

`=401.977 kev`


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