1.

The stopping potential for photoelectrons emitted from a surface illuminated by light wavelength of `5893A` is `036 V`. Calculate the maximum kinetic energy of photoelectrons, the work function of the surface, and the threshold frequency.

Answer» We know that
`KE_(max)=hf-phi((hc)/(lamda))-phi`
or `phi=(hc)/(lamda)-KE_(max)`
Also, `KE_(max)=ev_s=0.36eV`
`impliesphi=((6.62xx10^(34))xx(3xx10^(8)))/(5893xx10^(-10))-0.36xx1.6xx10^(-19)`
`=1.746eV`
The threshold frequency is given by
`f_0=(phi)/(h)=(2.794xx10^(-19))/(6.62xx10^(-34))=4.22xx10^(14)Hz`


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