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A radiation of energy E falls normally o...

A radiation of energy `E` falls normally on a perfctly refelecting surface . The momentum transferred to the surface is

A

`(2E)/(C^(2))`

B

`(E)/(C^(2))`

C

`(E)/(C )`

D

`(2E)/(C )`

Text Solution

Verified by Experts

The correct Answer is:
D
Energy of radiation, `E=hupsilon=(hC )/(lambda)`
Also, its momentum `p=(h)/(lambda)=(E)/(C )=p_(i)`
`p_(r)=-p_(i)=-(E)/(C )`
So, momentum transferred to the surface
`=p_(i)-p_(r)=(E)/(C )=(-(E)/(C ))=(2E)/(c)`
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