Question

The work function of Sodium is$2.75\mathrm{eV}$. What will be the Kinetic energy of emitted electron when the Photon of energy $3.54\mathrm{eV}$is incident on the surface of Sodium?

Answer 1

Kinetic Energy

• The photon energy incident on the metal surface is equal to the sum of the metal's work function and the electron's maximum kinetic energy.
• ${\mathrm{h}}_{\mathrm{f}}=\mathrm{K}.\mathrm{E}+\mathrm{\varphi }$

Where

• ${\mathrm{h}}_{\mathrm{f}}-$Photon energy
• $\mathrm{K}.\mathrm{E}-$Kinetic energy
• $\mathrm{\varphi }-$Work function

Then, Kinetic Energy is given by

• $\mathbf{K}\mathbf{.}\mathbf{E}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{h}}_{\mathbf{f}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{\varphi }$

Here given that

• $\mathrm{\varphi }=2.75\mathrm{eV}$
• ${\mathrm{h}}_{\mathrm{f}}=3.54\mathrm{eV}$

Now

• $\mathrm{K}.\mathrm{E}={\mathrm{h}}_{\mathrm{f}}-\mathrm{\varphi }$
• $\mathrm{K}.\mathrm{E}=3.54-2.75$
• $\mathrm{K}.\mathrm{E}=0.79\mathrm{eV}$

Therefore, the Kinetic energy of the emitted electron is $\mathbf{0}\mathbf{.}\mathbf{79}\mathbf{}\mathbf{eV}$