An electron and proton have same de-Broglie wavelength. Which one possess greater energy?

Options

A. zero
B. infinity
C. equal to K.E. of proton
D. greater than K.E. of proton

Answer

d

Explanation

K.E. of electron, `E_(1)=1/2 m_(1)v_(1)^(2)=1/2((m_(1)v_(1))^(2))/(m_(1))`
`=1/2(h^(2))/(lambda^(2)m_(1)) [ :' lambda=h/(m_(1)v_(1))]`
K.E. of photon, `E_(2)=1/2m_(2)v_(2)^(2)=1/2((m_(2)v_(2))^(2))/(m_(2))`
`=1/2(h^(2))/(lambda^(2)m_(2))`
`:. (E_(1))/(E_(2))=(m_(2))/(m_(1))gt1` or `E_(1) gt E_(2)`