The threshold wavelength for the photoelectri | vedclass

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Question

(C) The work function $\Phi_0$ is given by the formula $\Phi_0 = \frac{hc}{\lambda_0}$,where $h$ is Planck's constant $(6.63 	imes 10^{-34} \ J \cdot

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$4 	imes 10^{-19} \ J$

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(C) The work function $\Phi_0$ is given by the formula $\Phi_0 = \frac{hc}{\lambda_0}$,where $h$ is Planck's constant $(6.63 	imes 10^{-34} \ J \cdot s)$,$c$ is the speed of light $(3 	imes 10^8 \ m/s)$,and $\lambda_0$ is the threshold wavelength $(5000 \ Å = 5000 	imes 10^{-10} \ m = 5 	imes 10^{-7} \ m)$.Substituting the values:$\Phi_0 = \frac{(6.63 	imes 10^{-34}) 	imes (3 	imes 10^8)}{5 	imes 10^{-7}}$$\Phi_0 = \frac{19.89 	imes 10^{-26}}{5 	imes 10^{-7}}$$\Phi_0 = 3.978 	imes 10^{-19} \ J \approx 4 	imes 10^{-19} \ J$.Thus,the correct option is $C$.

The threshold wavelength for the photoelectric effect on sodium is $5000 \ Å$. Its work function is:

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