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A particle moving with kinetic energy E(...

A particle moving with kinetic energy `E_(1)` has de Broglie wavelength `lambda_(1)`. Another particle of same mass having kinetic energy `E_(2)` has de Broglie wavelength `lambda_(2). lambda_(2)= 3 lambda_(2)`. Then `E_(2) - E_(1)` is equal to

A

`7E_(1)`

B

`3E_(1)`

C

`8E_(1)`

D

`9E_(1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`p_(1)=h/(lamda_(1))impliesE_(1)=(p_(1)^(2))/(2m)impliesE_(1)=(h^(2))/(2mlamda_(1)^(2))impliesE_(2)=(h^(2))/(2mlamda_(2)^(2))=(9h^(2))/(2mlamda_(1)^(2))impliesE_(2)-E_(1)=(8h^(2))/(2mlamda_(1)^(2))=8E_(1)`
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