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The work function of a metal is 3.6 eV. What is the minimum frequency of incident radiation required to eject electrons from the surface of this metal?

Option: 1

1.5\times 10^{15}Hz


Option: 2

8.71\times 10^{14}Hz


Option: 3

6.0\times 10^{19}Hz


Option: 4

9.0\times 10^{20}Hz


Answers (1)

best_answer

The work function of the metal, \phi= 3.6eV.To eject an electron, the incident radiation must have energy equal to or greater than the work function. Using the relationship between energy and frequency of a photon,E= hv , we can calculate the minimum frequency required:
\phi= hv_{0}
v_{0}= \frac{\phi}{h}
v_{0}= \left ( 3.6eV \right )/\left ( 4.135 \times10^{-15}eVs \right )= 8.71\times10^{14}Hz.

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Suraj Bhandari

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