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An electron with initial kinetic energy ...

An electron with initial kinetic energy of `100 eV` is acclerated through a potential difference of `50 V`. Now the de-Brogile wavelength of electron becomes

A

(a)`1 Å`

B

(b)`sqrt(1.5Å)`

C

(c)`sqrt(3)Å`

D

(d)`12.27Å`

Text Solution

Verified by Experts

The correct Answer is:
A
For electron `lambda_(db)=sqrt(150/(100+50))=1 Å`
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