If the work function for the photoelectron em | vedclass

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(C) The work function $\phi$ is given by $\phi = 3.75 \ eV$. Converting to Joules: $\phi = 3.75 	imes 1.602 	imes 10^{-19} \ J \approx 6.0075 	imes

Vedclass · Accepted Answer

$330$

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(C) The work function $\phi$ is given by $\phi = 3.75 \ eV$. Converting to Joules: $\phi = 3.75 	imes 1.602 	imes 10^{-19} \ J \approx 6.0075 	imes 10^{-19} \ J$. The threshold wavelength $\lambda_0$ is calculated using the formula $\lambda_0 = \frac{hc}{\phi}$. Using $h = 6.626 	imes 10^{-34} \ J \cdot s$ and $c = 3 	imes 10^8 \ m/s$: $\lambda_0 = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{6.0075 	imes 10^{-19}} \approx 3.308 	imes 10^{-7} \ m$. Converting to nanometers: $\lambda_0 \approx 330.8 \ nm$. Thus,the approximate threshold wavelength is $330 \ nm$.

If the work function for the photoelectron emission of a metal is $3.75 \ eV$,then the threshold wavelength of the radiation needed for the ejection of the electron is approximately (in $nm$)

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