The wavelength of de - Broglie wave is `2 mu m` , then its momentum is `( h = 6.63 xx 10^(-34 J-s)`

Options

A. `3.315 xx10^(-28)` kg m/s
B. `1.66xx10^(-28)` kg m/s
C. `4.97xx10^(-28)` kg m/s
D. `9.9xx10^(-28)` kg m/s

Answer

A

Explanation

`lambda=h/p`
`:. p=h/lambda=(6.63xx10^(-34))/(2xx10^(-6))=3.3xx10^(-28)` kg m/s.