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The work functions for metals A, B and C are respectively 1.92 eV, 2.0 eV and 5 eV. According to Einstein's equation, the metals which will emit photo electrons for a radiation of wavelength 4100 Å is/are -
A
None
B
A only
C
A and B only
D
All the three metals
Detailed Solution
Photoelectrons are emitted only if the energy of the incident photon is greater than the work function of the metal.
Energy of a photon: $E = \dfrac{hc}{\lambda}$. A convenient form is $E\,(\text{eV}) = \dfrac{12400}{\lambda\,(\text{Å})}$
$E = \dfrac{12400}{4100} \approx 3$ eV
Metal A: work function 1.92 eV $\lt$ 3 eV, so A emits photoelectrons.
Metal B: work function 2.0 eV $\lt$ 3 eV, so B emits photoelectrons.
Metal C: work function 5 eV $\gt$ 3 eV, so C does not emit photoelectrons.
So the metals that will emit photoelectrons are A and B only.
Energy of a photon: $E = \dfrac{hc}{\lambda}$. A convenient form is $E\,(\text{eV}) = \dfrac{12400}{\lambda\,(\text{Å})}$
$E = \dfrac{12400}{4100} \approx 3$ eV
Metal A: work function 1.92 eV $\lt$ 3 eV, so A emits photoelectrons.
Metal B: work function 2.0 eV $\lt$ 3 eV, so B emits photoelectrons.
Metal C: work function 5 eV $\gt$ 3 eV, so C does not emit photoelectrons.
So the metals that will emit photoelectrons are A and B only.