Question:
The energy of a photon of wavelength 2500 is 4.96 eV. When photons of wavelength 3100 incident on a photosensitive material of work function 2.2 eV, the maximum velocity of the emitted photoelectrons is:
The energy of a photon of wavelength 2500 is 4.96 eV. When photons of wavelength 3100 incident on a photosensitive material of work function 2.2 eV, the maximum velocity of the emitted photoelectrons is:
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Always convert eV to Joules when using mass in kg!
Updated On: Jun 10, 2026
- $4 \times 10^6 ms^{-1}$
- $4 \times 10^5 ms^{-1}$
- $8 \times 10^6 ms^{-1}$
- $8 \times 10^5 ms^{-1}$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Einstein’s Photoelectric Equation: $K_{max} = E - \Phi$. $E = hc/\lambda$.
Step 2: Analysis
$E_{3100} = (4.96 \text{ eV} \times 2500) / 3100 = 4$ eV. $K_{max} = 4 \text{ eV} - 2.2 \text{ eV} = 1.8 \text{ eV} = 1.8 \times 1.6 \times 10^{-19}$ J $= 2.88 \times 10^{-19}$ J. $K_{max} = \frac{1}{2} m v^2 \implies v = \sqrt{2 K_{max} / m} = \sqrt{2 \times 2.88 \times 10^{-19} / 9 \times 10^{-31}} = \sqrt{0.64 \times 10^{12}} = 8 \times 10^5 ms^{-1}$.
Step 3: Conclusion
The max velocity is $8 \times 10^5 ms^{-1}$.
Final Answer: (D)
Einstein’s Photoelectric Equation: $K_{max} = E - \Phi$. $E = hc/\lambda$.
Step 2: Analysis
$E_{3100} = (4.96 \text{ eV} \times 2500) / 3100 = 4$ eV. $K_{max} = 4 \text{ eV} - 2.2 \text{ eV} = 1.8 \text{ eV} = 1.8 \times 1.6 \times 10^{-19}$ J $= 2.88 \times 10^{-19}$ J. $K_{max} = \frac{1}{2} m v^2 \implies v = \sqrt{2 K_{max} / m} = \sqrt{2 \times 2.88 \times 10^{-19} / 9 \times 10^{-31}} = \sqrt{0.64 \times 10^{12}} = 8 \times 10^5 ms^{-1}$.
Step 3: Conclusion
The max velocity is $8 \times 10^5 ms^{-1}$.
Final Answer: (D)
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