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An electron is accelerated under a potential difference of 64 V, the de-Brogile wavelength associated with electron is `[e = -1.6 xx 10^(-19) C, m_(e) = 9.1 xx 10^(-31)kg, h = 6.623 xx 10^(-34) Js]`

A

`1.53 A`

B

`2.53 A`

C

`3.35 A`

D

`4.54 A`

Text Solution

Verified by Experts

The correct Answer is:
A
Modified equation of de-Broglie wavelength:
`lambda = h/sqrt(2meV)`
`lambda = (12.27 A)/sqrt(V) = 12.27/sqrt(64) = 1.534 A`
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