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Radiation of energy 6.5 eV is incident ...

Radiation of energy 6.5 eV is incident on a metal surface whose work function is 4.2 eV. What is the potential difference that should be applied to stop the fastest photoelectrons emitted by the metal surface ?

A

1.3 V

B

2.3 V

C

3.5 V

D

5.5 V

Text Solution

Verified by Experts

The correct Answer is:
B
Max. K.E. `=hv-omega_(0)=6.5-4.2=2.3eV`
`:.` P.D.`=("Max. K.E.")/(e)=2.3V`
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MARVEL PUBLICATION-ELECTRONS AND PHOTONS -TEST YOUR GRASP - 17
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