The momentum of a particle having a de-Broglie wavelength of `10^(17)` m is (Given, `h=6.625xx10^(-34)m`)

Options

A. `3.3125xx10^(-7) kg ms^(-1)`
B. `26.5xx10^(-7) kg m s^(-1)`
C. `6.625xx10^(-17) kg ms^(-1)`
D. `13.25xx10^(-17) kg ms^(-1)`

Answer

C

Explanation

(c) According to de-Broglie relation ,
`lambda=(h)/(mv)=(h)/(p)`
where, `lambda`=wavelength
h=Planck's constant
p=momentum
Here, `h=6.625xx10^(-34) Js`
`lambda=10^(-17)m`
`:. P=(h)/(lambda)=(6.625xx10^(-34))/(10^(-17))`
`=6.625xx10^(-34)xx10^(17)`
`=6.625xx10^(-17) kg ms^(-1)`