Question

The work functions for several metals are shown in the table below. If 265 nm light is incident on each of these metals, determine: which metals exhibit the photoelectric effect.

	Element	Chemical Symbol	Work Function (eV)
1	Barium	Ba	2.48
2	Beryllium	Be	3.92
3	Cesium	Cs	1.92
4	Gold	Au	4.82
5	Lithium	Li	2.35
6	Mercury	Hg	4.53
7	Nickel	Ni	5.01

Group of answer choices

All the metals listed will exhibit the photoelectric effect.

All the metals listed except nickel will exhibit the photoelectric effect.

All the metals listed except nickel and gold will exhibit the photoelectric effect.

Of the metals listed only cesium, lithium, barium and beryllium will exhibit the photoelectric effect.

Of the metals listed only cesium, lithium and barium will exhibit the photoelectric effect.

Of the metals listed only cesium and lithium will exhibit the photoelectric effect.

Of the metals listed only cesium will exhibit the photoelectric effect.

Answer 1

The work function is the minimum energy required to remove electron, i.e, to show photoelectric effect.

Now the incident light wavelength, $\lambda=265nm$ which corresponds to energy, $E=\frac{hc}{\lambda}=\frac{6.626*10^{-34}*3*10^{8}}{265*10^{-9}}J=75.01*10^{-20}J=4.69eV$

So the incident light energy is less than the work function of Gold and Nickel.

So the right answer is: 3) All the metals listed except nickel and gold will exhibit the photoelectric effect.