The energy that should be added to an electron, to reduce its de-Broglie wavelength from `2xx10^(-9) m` to `0.5xx10^(-9)m` will be:

Options

A. 1.1 MeV
B. 0.56 MeV
C. 0.56KeV
D. 5.6 eV

Explanation

`lambda=h/p=h/(sqrt(2mE))`
`:. E=(h^(2))/(2m lambda^(2))`
`Delta E=(h^(2))/(2m)(1/(lambda_(1)^(2))-1/(lambda_(2)^(2)))`
Put `lambda_(1)=0/5xx10^(-9) m`
& `lambda_(2)=2xx10^(-9) m` and solve.