A metal has a work function of 2.5 × 10^ -19 J . When the metal is illuminated by a light beam of frequency 6.0 × 10^ 14 Hz , what is the stopping potential?

Question

Quantrex Academy · Accepted Answer

Correct answer: A. 0.91 V

The energy of the incident photon is given by

E = h\nu

.

Taking Planck's constant

h = 6.6 \times 10^{-34} \text{ J s}

:

E = (6.6 \times 10^{-34} \text{ J s}) \times (6.0 \times 10^{14} \text{ Hz}) = 3.96 \times 10^{-19} \text{ J}

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is:

K_{\text\max} = E - \Phi

K_{\text\max} = 3.96 \times 10^{-19} \text{ J} - 2.5 \times 10^{-19} \text{ J} = 1.46 \times 10^{-19} \text{ J}

The stopping potential

V_0

is related to the maximum kinetic energy by

K_{\text\max} = eV_0

.

V_0 = \dfrac{K_{\text\max}}{e} = \dfrac{1.46 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ C}} = 0.9125 \text{ V}

Rounding off, we get

V_0 \approx 0.91 \text{ V}

.

A metal has a work function of 2.5 × 10^ -19 J . When the metal is...

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