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

Answer

d

Explanation

As `lambda=h/(sqrt(2mE))` or `E=(h^(2))/(2mlambda^(2))`
`:. E_(1)=(h^(2))/(2mlambda_(1)^(2)) , E_(2)=(h^(2))/(2mlambda_(2)^(2))`
`:. E_(2)-E_(1)=(h^(2))/(2m)[1/(lambda_(2)^(2))-1/(lambda_(1)^(2))]`
`=((6.6xx10^(-34))^(2))/(2xx9.1xx10^(-31))[1/((0.5xx10^(-9))^(2))-1/((2xx10^(-9))^(2))]J`
`=((6.6xx10^(-34))^(2))/(2xx9.1xx10^(-31)xx10^(-18))xx3.75/(1.6xx10^(-19)) eV~~5.6 eV`