Practice portal › Dual Nature of Matter and Radiation › Einstein's Photoelectric Equation and Work Function

The photoelectric cut-off voltage in a certain experiment is 1.6 V. The maximum kinetic energy of the emitted photoelectrons is ______.

Asked in GSEB Board August 2020 · Speed and kinetic energy of the photoelectrons

• (1) 1.6 keV
• (2) 2.56 eV
• (3) 2.56×10⁻¹⁹ J
• (4) 1.6×10⁻¹⁹ J

Step-by-step solution

Why the other options are wrong

• (1) A cut-off of 1.6 V means Kₘₐₓ = eV₀ = 1.6 eV, not 1.6 keV; the kilo prefix makes it a thousand times too large.
• (2) Comes from multiplying 1.6 by 1.6 but keeping the unit eV. The factor 1.6×10⁻¹⁹ converts eV into joules, so 1.6 eV = 2.56×10⁻¹⁹ J, while 2.56 eV is a different, larger energy.
• (4) This is e×1 V, the energy for a 1 V cut-off. It forgets to multiply by V₀ = 1.6 V.

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