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The kinetic energy of an electron is 5 e...

The kinetic energy of an electron is `5 eV`. Calculate the de - Broglie wavelength associated with it `( h = 6.6 xx 10^(-34) Js , m_(e) = 9.1 xx 10^(-31) kg)`

A

`5.47 Å`

B

`109 Å`

C

`2.7 Å`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
`lambda = (h)/(sqrt(2 m E)) = (6.6 xx 10^(-34))/(sqrt(2 xx 9.1 xx 10^(-31) xx 5 xx 1.6 xx 10^(-19))`
`= 5.469 xx 10^(-10) m = 5.47 Å`
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