The energy that should be added to an electron to reduce its de Broglie wavelength from one nm to 0.5 nm is

Options

A. 2 times the initial kinetic energy
B. 3 times the initial kinetic energy
C. 0.5 times the initial kinetic energy
D. 4 times the initial kinetic energy

Answer

B

Explanation

de Broglie wavelength of an electron having kinetic energy K is
`lamda = (h)/(sqrt(2m_(e)K))` where `m_(e)` is the mass of an electron
`lamda^(2) = (h^(2))/(2m_(e)K) or K = (h)/(2m_(e) lamda^(2)), K prop (1)/(lamda^(2))`
`:. (K')/(K) = ((lamda)/(lamda')) = ((1nm)/(0.5nm))^(2) = (1)/(4)`
`K' = 4K`
Additional energy given `=K'-K = 4K -K = 3K`