Two monochromatic sources emit light at wavelengths \(\lambda\) and \(\lambda/2\). The stopping potentials for a photosensitive material using these two sources are found to be \(1\ \text{V}\) and \(3\ \text{V}\), respectively. What is the work function of the material?
Show Hint
Halving the wavelength doubles the energy of the incident photons.
Let \(E\) be the initial photon energy.
We have \(E - \phi = 1\ \text{eV}\) and \(2E - \phi = 3\ \text{eV}\).
Subtracting twice the first equation from the second equation directly yields \(\phi = 1\ \text{eV}\) with minimal calculation.
Step 1: Understanding the Question:
This question involves the photoelectric effect.
We need to determine the work function of a photosensitive material when illuminated by two different wavelengths, given their respective stopping potentials. Step 2: Key Formula or Approach:
• Einstein's Photoelectric Equation:
\[ e V_s = \frac{h c}{\lambda} - \phi \]
where \(V_s\) is the stopping potential, \(\phi\) is the work function, and \(\frac{h c}{\lambda}\) is the incident photon energy. Step 3: Detailed Explanation:
• For the first light source of wavelength \(\lambda\) and stopping potential \(1\ \text{V}\).:
\[ e (1) = \frac{h c}{\lambda} - \phi \implies 1\ \text{eV} = \frac{h c}{\lambda} - \phi \quad \text{--- (Eq. 1)} \]
• For the second light source of wavelength \(\lambda/2\) and stopping potential \(3\ \text{V}\).:
\[ e (3) = \frac{h c}{\lambda/2} - \phi \implies 3\ \text{eV} = 2 \left(\frac{h c}{\lambda}\right) - \phi \quad \text{--- (Eq. 2)} \]
• From Eq. 1, we can express the photon energy as:
\[ \frac{h c}{\lambda} = \phi + 1\ \text{eV} \]
• Substitute this expression into Eq. 2:
\[ 3\ \text{eV} = 2(\phi + 1\ \text{eV}) - \phi \]
\[ 3\ \text{eV} = 2\phi + 2\ \text{eV} - \phi \]
\[ 3\ \text{eV} = \phi + 2\ \text{eV} \]
\[ \phi = 3\ \text{eV} - 2\ \text{eV} = 1\ \text{eV} \]
Step 4: Final Answer:
The work function of the material is \(1\ \text{eV}\).
