Question

Find difference in work function of two different metal if their stopping potential are 0.4V and 1.6V respectively.(metals are illuminated with photons of same energy)

Hint:

When the incident light energy is constant, a higher stopping potential means the electrons had more kinetic energy, which implies they had to overcome a smaller work function. The difference in stopping potentials in Volts is numerically equal to the difference in work functions in electron-Volts (eV).

Answer 1

Step 1: Understanding the Concept:
Einstein's photoelectric equation relates the maximum kinetic energy of emitted photoelectrons to the energy of incident photons and the work function of the metal. The stopping potential is a direct measure of this maximum kinetic energy.

Step 2: Key Formula or Approach:
Einstein's photoelectric equation is:
\[ K_{\text{max}} = h\nu - \Phi \]
Where $K_{\text{max}} = eV_0$ ($V_0$ is stopping potential, $e$ is elementary charge).
So, $eV_0 = h\nu - \Phi$.

Step 3: Detailed Explanation:
Let the energy of the incident photons be $E = h\nu$.
For the first metal:
Stopping potential $V_{01} = 0.4$ V
Work function $\Phi_1 = E - e V_{01}$
For the second metal:
Stopping potential $V_{02} = 1.6$ V
Work function $\Phi_2 = E - e V_{02}$
We need to find the difference in their work functions, $\Delta\Phi = |\Phi_1 - \Phi_2|$:
\[ \Delta\Phi = |(E - e V_{01}) - (E - e V_{02})| \]
\[ \Delta\Phi = |e V_{02} - e V_{01}| = e |V_{02} - V_{01}| \]
Substitute the given values:
\[ \Delta\Phi = e |1.6 \text{ V} - 0.4 \text{ V}| \]
\[ \Delta\Phi = e (1.2 \text{ V}) = 1.2 \text{ eV} \]

Step 4: Final Answer:
The difference in work functions is 1.2 eV.