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According to de-Broglie hypothesis, t...

According to de-Broglie hypothesis, the wavelength associated with moving electron of mass 'm' is `'lambda_(e)'`. Using mass energy relation and Planck's quantum theory, the wavelength associated with photon is `'lambda_(p)'`. If the energy (E) of electron and photonm is same, then relation between `lambda_e` and `'lambda_(p)'` is

A

`lambda_(p)alpha lambda_(e)`

B

`lambda_(p)alpha lambda_(e)^(2)`

C

`lambda_(p)alpha sqrt(lambda_(e))`

D

`lambda_(p)alpha 1/sqrt(lambda_(e))`

Text Solution

Verified by Experts

The correct Answer is:
A
For photon: `E=(hc)/(lambda_(p))`
`therefore lambda_(p)=(hc)/(E)......(i)`
For electron `E=mc^(2)=pc`
`therefore p=E/C`
`lambda_(e)=h/p=(hc)/(E).....(ii)`
By eq. (i) and (ii), `lambda_(p) alpha lambda_(e)`
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