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A metal surface has a work function of 4.5 eV. What is the minimum frequency of incident radiation required for photoelectric emission to occur?

Option: 1

1.0\times 10^{14}Hz


Option: 2

1.08 \times 10^{15}Hz


Option: 3

3.7\times 10^{16}Hz


Option: 4

5.8\times 10^{17}Hz


Answers (1)

best_answer

The minimum frequency of incident radiation required for photoelectric emission is given by the equation

 \phi= hv_{0} ,

where \phi is the work function, h is Planck's constant and v_{0} is the minimum frequency of incident radiation required.

Substituting the given values, we get  v_{0}= \phi/h= \left ( 4.5eV \right )/\left ( 6.626\times 10^{-34}Js \right )= 1.08 \times10^{15}Hz.

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manish painkra

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