If uncertainity in position is zero, the uncertainity in momentum of an electron will be

Options

A. `1/m sqrt(h/(4lambda))`
B. `msqrt(h)/(sqrt(4lambda))`
C. `sqrt((h)/(4lambdam))`
D. `sqrt((hm)/(4lambda))`

Answer

D

Explanation

`Delta x . ""m Delta v ge h/(4pi)` and `lambda = h/(mv)`
In the present case `Delta x = Delta v `
`therefore m (Deltav)^2 ge h/(4pi) implies m^2 (Deltav)^2 ge (mh)/(4pi)`
`implies m.Delta x ge sqrt((mh)/(4pi))`