Question

A potassium metal photoelectric surface has a work function Φ = 2.24 eV.

a) Find the maximum kinetic energy of the electrons emitted when the surface is illuminated with light having frequency 6.00 x 1015 Hz. (ANS: 22.6 eV or 3.62 x 10-18 J)

b) Find the stopping potential for these electrons. (ANS: 22.6 V)

Answer 1

Answer on Question #43162-Physics-Nuclear Physics

A potassium metal photoelectric surface has a work function $\Phi = 2.24\,\mathrm{eV}$.

a) Find the maximum kinetic energy of the electrons emitted when the surface is illuminated with light having frequency $6.00 \times 10^{15}\,\mathrm{Hz}$. (ANS: 22.6 eV or 3.62 x 10-18 J)

b) Find the stopping potential for these electrons. (ANS: 22.6 V)

Solution

a) the maximum kinetic energy of the electrons is

$$K_{\max} = h f - \Phi = (4.14 \cdot 10^{-15}\,\mathrm{eV} \cdot \mathrm{s}) \cdot (6.00 \cdot 10^{15}\,\mathrm{Hz}) - 2.24\,\mathrm{eV} = 22.6\,\mathrm{eV} \text{ or } 3.62 \cdot 10^{-18}\,\mathrm{J}.$$

b) the stopping potential for these electrons is

$$V_{stop} = \frac{K_{\max}}{e} = \frac{22.6\,\mathrm{eV}}{1\,\mathrm{e}} = 22.6\,\mathrm{V}.$$

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