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What is the work function of a metal whose threshold frequency is 1.5\times10^{15}Hz?

Option: 1

 4.72 eV


Option: 2

 7.38 eV


Option: 3

9.45 eV


Option: 4

12.89 eV


Answers (1)

best_answer

The work function of a metal is given by the equation

 \phi= hv_{0}= hc/\lambda _{0},

where h is Planck's constant, c is the speed of light, and \lambda _{0}  is the threshold wavelength. The threshold frequency is given by the equation v_{0}= c/\lambda _{0}. Substituting the given value, we get v_{0}= 1.5\times 10^{15}Hz.
Therefore, the work function can be calculated as
\phi=hc/\lambda _{0}= \left ( 6.626\times 10^{-34}Js \right )\left ( 3\times 10^{8}m/s \right )/\left ( 1.5\times 10^{15}Hz \right )= 4.72eV
 

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