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If v is frequency, lambda is the wavelen...

If v is frequency, `lambda` is the wavelength and `barv` is the wave number then the energy of a photon can be represented by

A

`hv`

B

`hcbar(v)`

C

`hclambda`

D

`hc//lambda`

Text Solution

Verified by Experts

The correct Answer is:
a,b,d
Wavelength `barv=1//lambda`,
Energy of photon `=hv=(hc)/lambda=hcbar(v)`
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