The energy that should be added to an electron, to reduce its de-Broglie wavelength from `2 xx 10^(-9)m` to `0.5 xx 10^(-9)m` will be:

Options

A. `1.1 MeV`
B. `0.56 MeV`
C. `0.56 KeV`
D. `5.67eV`

Answer

D

Explanation

`lambda = (h)/(p) = (h)/(sqrt(2mE)) :. E = (h^(2))/(2m lambda^(2))`
`DeltaE = (h^(2))/(2m) ((1)/(lambda_(1)^(2)) - (1)/(lambda_(2)^(2)))`
Put `lambda_(1) = 0.5 xx 10^(-9)m` and `lambda_(2) = 2 xx 10^(-9)m` and solve.