Light of wavelength 3500 mathring A is incid | vedclass

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(B) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Substituting the values: $E = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 1

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Metal $B$

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(B) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Substituting the values: $E = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{3500 	imes 10^{-10}} \ J$.Converting to electron-volts $(eV)$: $E = \frac{12400 \ \mathring A \cdot eV}{3500 \ \mathring A} \approx 3.54 \ eV$.Photoelectric emission occurs if the incident energy $E$ is greater than the work function $\phi$ of the metal $(E > \phi)$.For metal $A$: $\phi_A = 4.2 \ eV$. Since $3.54 \ eV For metal $B$: $\phi_B = 1.9 \ eV$. Since $3.54 \ eV > 1.9 \ eV$,metal $B$ will emit photoelectrons.Therefore,only metal $B$ will emit photoelectrons.

Light of wavelength $3500 \ \mathring A$ is incident on two metals $A$ and $B$. If their work functions are $4.2 \ eV$ and $1.9 \ eV$ respectively,which metal will emit photoelectrons?

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