The momentum of a particle which has a de Broglie wavelength of `2.5 xx 10^-10 m` is.

Options

A. `2.64 xx 10^-24 kg m sec^-1`
B. `3.62 xx 10^-24 kg m sec^-1`
C. `4.64 xx 10^-24 kg m sec^-1`
D. `3.62 xx 10^-26 kg m sec^-1`

Answer

A

Explanation

(a) `h = 6.6 xx 10^-34 kg m^2 s^-1`
Momentum `= (h)/(lamda)` (using de Borglie equation)
=`(6.6 xx 10^-34)/(2.5 xx 10^-10) = 2.64 xx 10^-24 kg m sec`.