What should be the percentage increase or decrease of kinetic energy of an electron so that its de Broglie wavelength is halved?

Explanation

Kinetic energy, `E = (p^(2))/(2m)` [p = momentum]
`:. p = sqrt(2mE)`
de Broglie wavelength, `lamda = (h)/(p) = (h)/(sqrt(2mE))`
`:. (lamda_(1))/(lamda_(2)) = sqrt((E_(2))/(E_(1)))`
Here, `lamda_(2) = (lamda_(1))/(2) or, (lamda_(1))/(lamda_(2)) = 2`
Hence, `sqrt((E_(2))/(E_(1))) = 2 or, E_(2) = 4E_(1)`
`:.` Increase in kinetic energy
`= (E_(2) - E_(1))/(E_(1)) xx 100% = (4E_(1) - E_(1))/(E_(1)) xx 100% = 300%`