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(B) The photoelectric effect occurs when the energy of incident photons $E = \frac{hc}{\lambda}$ is greater than or equal to the work function $\phi$

Vedclass · Accepted Answer

$650$

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(B) The photoelectric effect occurs when the energy of incident photons $E = \frac{hc}{\lambda}$ is greater than or equal to the work function $\phi$ of the metal.Given the range of work function $2 eV \leq \phi \leq 5 eV$, the threshold wavelength $\lambda$ must satisfy $\lambda \leq \frac{hc}{\phi}$.For $\phi = 2 eV$, $\lambda_{\max} = \frac{4 	imes 10^{-15} eVs 	imes 3 	imes 10^{8} m/s}{2 eV} = 6 	imes 10^{-7} m = 600 nm$.For $\phi = 5 eV$, $\lambda_{\min} = \frac{4 	imes 10^{-15} eVs 	imes 3 	imes 10^{8} m/s}{5 eV} = 2.4 	imes 10^{-7} m = 240 nm$.Thus, the range of wavelengths that can cause the photoelectric effect is $240 nm \leq \lambda \leq 600 nm$.Comparing the given options, $650 nm$ is greater than $600 nm$, so it cannot cause the photoelectric effect.

The work function of metals is in the range of $2 eV$ to $5 eV$. Find which of the following wavelengths of light cannot be used for the photoelectric effect (in $nm$)? (Consider, Planck constant $= 4 \times 10^{-15} eVs$, velocity of light $= 3 \times 10^{8} m/s$)

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