1.

The wavelength of de-Broglie wave is 3 mu m, then its momentum (h = 6.63 xx 10^(-34) Js) is

Answer»

`3.315 XX 10^(-28) "kg ms"^(-1)`
`2.21 xx 10^(-28) "kg ms"^(-1)`
`4.97 xx 10^(-28) "kg ms"^(-1)`
`9.9 xx 10^(-28) "kg ms"^(-1)`

Solution :`p = (h)/(lambda) = (6.03 xx 10^(-34) JS)/(2 xx 10^(-6) m) PROP 3.315 xx 10^(-28) "kg ms"^(-1)`


Discussion

No Comment Found

Related InterviewSolutions