© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Note the Given Information
The metal surface is illuminated by two different photon energies:
• E_1 = 4\, \text{eV}
• E_2 = 2.5\, \text{eV}
The ratio of the maximum speeds of the emitted photoelectrons for these two energies is given by:
\frac{V_{1,\max}}{V_{2,\max}} = 2.
We denote the work function of the metal by \phi_0 .
Step 2: Express the Kinetic Energy in Terms of Photon Energy and Work Function
The maximum kinetic energy of the photoelectrons in each case is:
K_{\max} = E - \phi_0,
where E is the incident photon energy and \phi_0 is the work function of the metal.
Hence, for each set of incident photons, we have:
K_{1,\max} = E_1 - \phi_0 = 4 - \phi_0 \,\text{(eV)},
K_{2,\max} = E_2 - \phi_0 = 2.5 - \phi_0 \,\text{(eV)}.
Step 3: Relate Kinetic Energy to Velocity
The maximum kinetic energy of the emitted electrons is also given by:
K_{\max} = \frac{1}{2} \, m \, (V_{\max})^2.
Therefore, the ratio of kinetic energies for the two cases becomes:
\frac{K_{1,\max}}{K_{2,\max}}
= \frac{\frac{1}{2}m (V_{1,\max})^2}{\frac{1}{2}m (V_{2,\max})^2}
= \frac{(V_{1,\max})^2}{(V_{2,\max})^2}.
Step 4: Insert the Known Velocity Ratio
We know \frac{V_{1,\max}}{V_{2,\max}} = 2 , so
\frac{(V_{1,\max})^2}{(V_{2,\max})^2}
= \left(\frac{V_{1,\max}}{V_{2,\max}}\right)^2
= 2^2 = 4.
On the other hand, from the difference in photon energies:
\frac{K_{1,\max}}{K_{2,\max}}
= \frac{E_1 - \phi_0}{E_2 - \phi_0}
= \frac{4 - \phi_0}{2.5 - \phi_0}.
Equating these two expressions gives:
\frac{4 - \phi_0}{2.5 - \phi_0} = 4.
Step 5: Solve for the Work Function \phi_0
Cross-multiply to find \phi_0 :
4 - \phi_0 = 4 \times (2.5 - \phi_0).
Expand the right-hand side:
4 - \phi_0 = 10 - 4\phi_0.
Rearrange the equation:
10 - 4\phi_0 = 4 - \phi_0
\quad \Rightarrow \quad
10 - 4 = 4\phi_0 - \phi_0
\quad \Rightarrow \quad
6 = 3\phi_0
\quad \Rightarrow \quad
\phi_0 = 2 \,\text{eV}.
Final Answer
The work function of the metal is 2\,\text{eV} .
