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The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted?

Asked in GSEB Board July 2023 · Speed and kinetic energy of the photoelectrons

• (1) Zero
• (2) 9×10⁻¹⁹ J
• (3) 2.4×10⁻¹⁹ J
• (4) 1.0×10⁻¹⁹ J

Step-by-step solution

Why the other options are wrong

• (1) A cut-off voltage of 1.5 V is needed precisely because the fastest electrons do have kinetic energy. Zero energy would mean a cut-off voltage of zero.
• (2) No single slip gives 9×10⁻¹⁹ J. Sanity check: Kₘₐₓ = eV₀ = 1.6×10⁻¹⁹×1.5 = 2.4×10⁻¹⁹ J.
• (4) No single slip gives 1.0×10⁻¹⁹ J. Sanity check: Kₘₐₓ = eV₀ must equal 1.5 eV, and 1.5×1.6×10⁻¹⁹ = 2.4×10⁻¹⁹ J.

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