A monochromatic light source with an intensity of 5 mW emits 8 × 10^ 15 photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this arrangement is 2.0 V . Determine the

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Correct answer: B. 1.9 eV

The energy of a single photon is given by

E = \dfrac{P}{n}

Substituting the given values:

E = \dfrac{5 \times 10^{-3}}{8 \times 10^{15}} = 6.25 \times 10^{-19} \text{ J}

Converting the energy into electron-volts:

E = \dfrac{6.25 \times 10^{-19}}{1.6 \times 10^{-19}} \text{ eV} \approx 3.9 \text{ eV}

From Einstein's photoelectric equation:

E = \Phi + K_\max

Since the stopping potential is

V_s = 2.0 \text{ V}

, the maximum kinetic energy is

K_\max = 2.0 \text{ eV}

.

\Phi = E - K_\max = 3.9 \text{ eV} - 2.0 \text{ eV} = 1.9 \text{ eV}

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