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

b. AG, work function=4.74eV

Step-by-step 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=(hc)/(\lambda)`

where

h is the Planck constant

c is the speed of light

Substituting,

`E=((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=(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.