The de Broglie wavelength of an electron is `0.4 xx 10^(-10) m` when its kinetic energy is 1.0 keV. Its wavelength will be `1.0 xx 10^(-10) m`, when its kinetic energy is

Options

A. 0.2 keV
B. 0.8 keV
C. 0.63 keV
D. 0.16 keV

Answer

D

Explanation

de Broglie wavelength, `lamda = (h)/(p) = (h)/(sqrt(2m E_(k)))` [ p = momentum, `E_(k)`= kinetic energy]
`:. (lamda_(1))/(lamda_(2)) = sqrt(((E_(k))_(2))/((E_(k))_(1)))`
`:. (E_(k))_(2) = ((lamda_(1))/(lamda_(2)))^(2) xx (E_(k))_(1) = ((0.4 xx 10^(-10))/(1.0 xx 10^(-10)))^(2) xx 1`
= 0.16 keV