Question

Photons with energy ${\mathrm{E}}_1=4\mathrm{eV}$ and ${\mathrm{E}}_2=2.5\mathrm{eV}$ alternately illuminate the surface of a metal. The ratio of the maximum speeds of the photoelectrons released in the two scenarios is $2$. The metal's work function in electron volts is $\_\_\_\_$.

Answer 1

Einstein's photoelectric equation

• According to Einstein's theory, an electron receives the complete energy of a photon that strikes a metal surface.
• $\mathrm{Kinetic}\mathrm{energy}=\mathrm{Energy}\mathrm{of}\mathrm{Photon}-\mathrm{Work}\mathrm{Function}$
• Let ${\mathrm{v}}_1$ and ${\mathrm{v}}_2$ represent the photo-electrons' velocities and let ${\mathrm{\varphi }}_0$ represent the work function of the metal.
• ${\mathrm{E}}_1=4\mathrm{eV}$
• ${\mathrm{E}}_2=2.5\mathrm{eV}$
• Using the photoelectric equation proposed by Einstein,

In First case: $\frac{1}{2}{{\mathrm{mv}}_1}^2=4-{\mathrm{\varphi }}_0\cdots \left(\mathrm{i}\right)$

In Second case: $\frac{1}{2}{{\mathrm{mv}}_2}^2=2.5-{\mathrm{\varphi }}_0\cdots \left(\mathrm{ii}\right)$

It is given that: $\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=2$

Now,

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{1}{2}}{{\mathrm{mv}}_1}^2}{\frac{1}{2}{{\mathrm{mv}}_2}^2}=\frac{4-{\mathrm{\varphi }}_0}{2.5-{\mathrm{\varphi }}_0} \\ \Rightarrow {\left(2\right)}^2=\frac{4-{\mathrm{\varphi }}_0}{2.5-{\mathrm{\varphi }}_0} \\ \Rightarrow 10-4{\mathrm{\varphi }}_0=4-{\mathrm{\varphi }}_0 \\ \Rightarrow {\mathrm{\varphi }}_0=2\mathrm{eV} \end{gathered}$$

Therefore, The metal's work function in electron volts is $2\mathrm{eV}$.