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What is the work function of a metal whose threshold frequency is 5.5\times 10^{14}Hz ?

Option: 1

3.41 eV


Option: 2

2.06 eV


Option: 3

1.24 eV


Option: 4

8.60 eV


Answers (1)

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Work function \left ( \phi \right ) is given by \phi= hv_{0}= hc/\lambda _{0}, where h is Planck's constant, c is the speed of light, v_{0}  is the threshold frequency, and \lambda _{0} is the threshold wavelength. Substituting the given values, we get
\phi= \left ( 6.626\times 10^{-34}Js \right )\left ( 3\times 10^{8}m/s \right )/\left ( 5.5\times 10^{14}Hz \right )= 2.06eV.

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himanshu.meshram

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