Kinetic energy of a free electron is doubled. By how many times, would its de Broglie wavelength increase?

Explanation

Kinetic energy, `E = (p^(2))/(2m)` [p = monmentum of electron]
`:. p= sqrt(2mE)`
de Broglie wavelength,
`lamda = (h)/(p) = (h)/(sqrt(2mE))`
i.e., `(lamda_(1))/(lamda_(2)) = sqrt((E_(2))/(E_(1))) or, lamda_(2) = lamda_(1) sqrt((E_(1))/(E_(2))) = lamda_(1) sqrt((1)/(2)) = (lamda_(1))/(sqrt2)`
Hence, de Broglie wavelength will be `(1)/(sqrt2)` times the original one