By what factor does the de-Broglie wavelength of a free electron changes if its kinetic energy is doubled ?

Options

A. `1/2`
B. `2`
C. `(1)/(sqrt2)`
D. `sqrt2`

Answer

C

Explanation

We know
`lamda = ( h)/(mv)`
and `K = (1)/(2) mv ^(2) = (( mv ) ^(2))/( 2m )`
`mv = sqrt (2 mK )`
Thus, `lamda = (h )/( sqrt (2 mK))`
`implies lamda `
`prop (1)/(sqrt K)`
` therefore (lamda _(2))/( lamda _(1)) = ( sqrt K _(1))/( sqrtK _(2))`
`implies (sqrt K _(1))/( sqrt(2K _(1))) (because K _(2) = 2 K _(1))`
`implies (lamda _(2))/( lamda _(1)) = (1)/( sqrt2)`
` therefore lamda = (lamda _(1))/( sqrt2)`