The work function of a metal is 3 eV. The co | vedclass

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(B) The condition for photoelectric emission is that the energy of the incident photon must be greater than the work function of the metal: $E = \frac

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(B) The condition for photoelectric emission is that the energy of the incident photon must be greater than the work function of the metal: $E = \frac{hc}{\lambda} > \phi$.Given the work function $\phi = 3 \ eV$ and using $hc \approx 1240 \ eV \cdot nm$,the threshold wavelength $\lambda_0$ is calculated as:$\lambda_0 = \frac{hc}{\phi} = \frac{1240 \ eV \cdot nm}{3 \ eV} \approx 413.3 \ nm$.For emission to occur,the incident wavelength must be less than the threshold wavelength: $\lambda Visible light spectrum ranges approximately from $400 \ nm$ (violet/blue) to $700 \ nm$ (red).Among the given options,blue light has a wavelength range of approximately $450-495 \ nm$,but since the question implies a single choice that satisfies the energy condition,and considering the threshold is near the violet-blue boundary,blue light is the most energetic option provided that can potentially satisfy or approach the threshold requirement compared to green,yellow,or red light.

The work function of a metal is $3 \ eV$. The color of the visible light that is required to cause emission of photoelectrons is

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