Question

Which of the following metals require ultraviolet light to exhibit the photoelectric effect?The options available: a. Cs, work function=1.95eV b. AG, work function=4.74eV c. K, work function=2.29eV d. Y, work function=3.10eV

Answer 1

Answer:

b. AG, work function=4.74eV

Explanation:

Ultraviolet light starts at the end of the visible light spectrum, where violet light ends:

$\lambda=380 nm =3.8\cdot 10^{-7}m$ (wavelength of lowest-energy ultraviolet light)

So, the lowest energy of ultraviolet light can be found by using the formula

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

Substituting,

$E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.8\cdot 10^{-7} m}=5.23\cdot 10^{-19}J$

And keeping in mind that

$1 eV = 1.6\cdot 10^{-19}J$

This energy converted into electronvolts is

$E=\frac{5.23\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=3.27 eV$

The work function of a metal is the minimum energy needed to extract a photoelectron from the surface of the metal. Therefore, the metals that exhibit photoelectric effect are the ones whose work function is larger than the energy we found previously, so:

b. AG, work function=4.74eV

Because for all the other metals, visible light will be enough to extract photoelectrons.