Question

Light of wavelength $488\,nm$ is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is $0.38V$ . Find the work function (in eV) of the material from which the emitter is made.

Answer 1

To obtain the work function (in eV) of the material from which the emitter is created, we will apply the formula for the work function of the material from which the emitter is made from Einstein's photoelectric effect and solve for the query.

Formula used:
${\phi _0}\; = {\text{ }}\dfrac{{hc}}{\lambda } - e{V_0}$

Complete step by step answer:
The least amount of energy necessary to start the emission of electrons from a metal's surface is known as its work function. Now, coming to the question, in order to answer the given question, let us first write all the values that we are provided within the question: The argon laser produces a wavelength of light that is,
$\lambda = 488{\text{ }}nm = 488 \times {10^{ - 9}}\;m$
Photoelectrons' potential for stopping, ${V_0}\; = {\text{ }}0.38{\text{ }}V$
And we know that, Planck’s constant, $\left( h \right)$ is given by $h = 6.6 \times {10^{ - 34}}\;Js$
Charge on an electron, $e{\text{ }} = 1.6{\text{ }} \times {10^{ - 19}}\;C$
Light's speed, $c = 3 \times {10^8}\;m/s$

We have the relationship involving the work function ${\phi _0}$ of the emitter's material as: $e{V_0} = {\text{ }}\dfrac{{hc}}{\lambda } - {\phi _0}$ from Einstein's photoelectric effect.Therefore, from this equation we will find out the value for ${\phi _0}$.
${\text{ }}{\phi _0}\; = {\text{ }}\dfrac{{hc}}{\lambda } - e{V_0}$
Now, we will plug in all the known values in the above equation;
${\text{ }}{\phi _0} = \left( {\dfrac{{6.6{\text{ }}x{\text{ }}{{10}^{ - 34}}\; \times {\text{ }}3{\text{ }} \times {\text{ }}{{10}^8}}}{{1.6{\text{ }}x{\text{ }}{{10}^{ - 19}}\; \times {\text{ }}488{\text{ }} \times {\text{ }}{{10}^{ - 9}}}}} \right) - \left( {\dfrac{{1.6{\text{ }} \times {{10}^{ - 19}}\; \times {\text{ }}0.38}}{{1.6{\text{ }} \times {\text{ }}{{10}^{ - 19}}}}} \right)$
$\Rightarrow {\text{ }}{\phi _0} = 2.54 - 0.38{\text{ }} \\\ \therefore {\text{ }}{\phi _0}= 2.16{\text{ }}eV$

As a result, the material used to make the emitter has a work function of $2.16{\text{ }}eV$.

Note: When metals with differing work functions are combined, electrons tend to leave the lower work function metal (where they are less firmly bound) and migrate to the higher work function metal. When making connections between dissimilar metals in specific electronic circuits, this impact must be considered. Because some electrons in a material are held more strongly than others, a precise work function definition defines which electrons are involved, which are normally the most loosely bound.