Question:
The work function of a metal M is 6.3 eV. The wavelength of the incident radiation required to just eject the electrons from its surface (in nm) is:
The work function of a metal M is 6.3 eV. The wavelength of the incident radiation required to just eject the electrons from its surface (in nm) is:
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Use the relation \(E = \frac{hc}{\lambda}\) to find the threshold wavelength for photoemission. Remember to convert eV to joules when using SI units.
Updated On: Jun 26, 2026
- 102
- 330
- 197
- 310
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The Correct Option is C
Solution and Explanation
Step 1: Recall the photoelectric effect formula.
The energy of a photon is given by \[ E = \frac{hc}{\lambda} \] where \(E\) is the work function (in eV), \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength.
Step 2: Convert energy to SI units.
The work function \( \phi = 6.3 \, \text{eV} \). In joules, \[ \phi = 6.3 \times 1.602 \times 10^{-19} \, \text{J} = 1.009 \times 10^{-18} \, \text{J}. \]
Step 3: Calculate wavelength.
\[ \lambda = \frac{hc}{\phi} = \frac{6.626\times 10^{-34} \times 3 \times 10^8}{1.009 \times 10^{-18}} \approx 197 \times 10^{-9} \, \text{m} = 197 \, \text{nm} \]
Step 4: Conclusion.
The required wavelength of the incident radiation is \[ \lambda \approx 197 \, \text{nm}. \]
The energy of a photon is given by \[ E = \frac{hc}{\lambda} \] where \(E\) is the work function (in eV), \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength.
Step 2: Convert energy to SI units.
The work function \( \phi = 6.3 \, \text{eV} \). In joules, \[ \phi = 6.3 \times 1.602 \times 10^{-19} \, \text{J} = 1.009 \times 10^{-18} \, \text{J}. \]
Step 3: Calculate wavelength.
\[ \lambda = \frac{hc}{\phi} = \frac{6.626\times 10^{-34} \times 3 \times 10^8}{1.009 \times 10^{-18}} \approx 197 \times 10^{-9} \, \text{m} = 197 \, \text{nm} \]
Step 4: Conclusion.
The required wavelength of the incident radiation is \[ \lambda \approx 197 \, \text{nm}. \]
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