Question:
For a photosensitive material, work function is '\(W_0\)' and stopping potential is 'V'. The wavelength of the incident radiation is (h=Planck's constant, c = velocity of light, e = electronic charge)
For a photosensitive material, work function is '\(W_0\)' and stopping potential is 'V'. The wavelength of the incident radiation is (h=Planck's constant, c = velocity of light, e = electronic charge)
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Einstein's equation: photon energy equals work function plus maximum kinetic energy.
Updated On: Oct 1, 2026
- \(\frac{\text{hc}}{(\text{W}_0-\text{eV})}\)
- \(\frac{\text{hc}}{(\text{W}_0+\text{eV})}\)
- \(\text{hc}(\text{W}_0-\text{eV})\)
- \(\text{hc}(\text{W}_0+\text{eV})\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Einstein's photoelectric equation: \(h\nu = W_0 + K_{max}\). The maximum kinetic energy is \(eV\), where \(V\) is the stopping potential.
Step 2: Write the equation:
\[ \frac{hc}{\lambda} = W_0 + eV \]
Step 3: Solve for lambda:
\[ \lambda = \frac{hc}{W_0 + eV} \]
Step 4: Check:
Option (B). Option (A) has a minus sign, which would apply if the radiation had less energy than the work function plus kinetic energy. Options (C) and (D) multiply by \(hc\) instead of dividing, which gives the wrong units for a wavelength.
Final Answer:
Photon energy = W0 + eV, so lambda = hc / (W0 + eV). \[ \boxed{\text{(B) }\dfrac{hc}{W_0+eV}} \]
Einstein's photoelectric equation: \(h\nu = W_0 + K_{max}\). The maximum kinetic energy is \(eV\), where \(V\) is the stopping potential.
Step 2: Write the equation:
\[ \frac{hc}{\lambda} = W_0 + eV \]
Step 3: Solve for lambda:
\[ \lambda = \frac{hc}{W_0 + eV} \]
Step 4: Check:
Option (B). Option (A) has a minus sign, which would apply if the radiation had less energy than the work function plus kinetic energy. Options (C) and (D) multiply by \(hc\) instead of dividing, which gives the wrong units for a wavelength.
Final Answer:
Photon energy = W0 + eV, so lambda = hc / (W0 + eV). \[ \boxed{\text{(B) }\dfrac{hc}{W_0+eV}} \]
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