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 photoelectrons for a radiation of wavelength 4100 Å is/are

Options

A. A only
B. A and B only
C. All the three metals
D. C only

Answer

B

Explanation

Energy of the incident radiation `=(hc)/(lambda)`
`=(6.63xx10^(-34)xx3xx10^(8))/(41xx10^(-8))J`
`=(19.89xx10^(-18))/(41xx1.6xx10^(-19))eV=3eV`
For metals A and B, `W_(A)=1.92 eV` and `W_(B)=2eV`
`:.` This is less than 3 eV, the energy of incident radiation.
`:.` Photoemission is possible for A and B.
However for C, `W_(C)=5eV`. Hence the photoemission is not possible.