A polished copper sphere of radius 5.0 cm hangs on an insulating thread in a large vacuum chamber…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A polished copper sphere of radius $5.0\ \text{cm}$ hangs on an insulating thread in a large vacuum chamber, far from the walls. At first it carries no charge. Ultraviolet light of wavelength $200\ \text{nm}$ from a lamp outside shines on it through a quartz window. The work function of copper is $4.65\ \text{eV}$ . The light delivers $2.0\ \mu\text{W}$ to the sphere, and on average one photon in $10^{4}$ frees an electron. Take $h = 6.63 \times 10^{-34}\ \text{J s}$ , $c = 3.0 \times 10^{8}\ \text{m/s}$ , $e = 1.6 \times 10^{-19}\ \text{C}$ and $\dfrac{1}{4\pi\varepsilon_0} = 9.0 \times 10^{9}\ \text{N m}^2\text{/C}^2$ .

The freed electrons leave with a spread of energies, up to a largest value. As they leave, the sphere becomes positively charged and starts to pull them back. In the end no electron can escape any more, and the charge stops growing.

What is the energy of one photon, in electronvolts?
What is the largest kinetic energy of the freed electrons?
What is the final potential of the sphere?
What is the final charge on the sphere?
How many electrons has the sphere lost by then?
At first every freed electron escapes. Using the starting rate of escape, find the least time the sphere needs to reach its final charge.
The experiment is repeated with a copper sphere of radius $10.0\ \text{cm}$ . What is its final charge?

A student says a larger sphere would charge to a higher potential, because it catches more light. Check this against your answers.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. The photoelectric equation gives the fastest electron. The sphere stops charging when its potential holds back even that electron. Its capacitance then gives the charge.

Step 1. Photon energy (a). $E = \dfrac{hc}{\lambda} = \dfrac{6.63 \times 10^{-34}\ \text{J s} \times 3.0 \times 10^{8}\ \text{m/s}}{200 \times 10^{-9}\ \text{m}} = 9.945 \times 10^{-19}\ \text{J} = 6.22\ \text{eV}$ .

Step 2. Fastest electron (b). $K_{\max} = 6.22\ \text{eV} - 4.65\ \text{eV} = 1.57\ \text{eV}$ . (The light is well above copper's threshold, $\lambda_0 = hc/\phi = 267\ \text{nm}$ .)

Step 3. Final potential (c). An electron leaving the surface with $K_{\max}$ just fails to escape when $eV = K_{\max}$ . So $V = 1.57\ \text{V}$ .

Step 4. Charge (d, e). $Q = \dfrac{RV}{k} = \dfrac{0.05\ \text{m} \times 1.566\ \text{V}}{9.0 \times 10^{9}\ \text{N m}^2\text{/C}^2} = 8.70\ \text{pC}$ . That is $Q/e = 5.44 \times 10^{7}$ electrons.

Step 5. Least time (f). The lamp delivers $\dfrac{2.0 \times 10^{-6}\ \text{W}}{9.945 \times 10^{-19}\ \text{J}} = 2.01 \times 10^{12}$ photons per second, so $2.01 \times 10^{8}$ electrons leave per second at first (a current of $32.2\ \text{pA}$ ). Later, slow electrons are pulled back, so the real time is longer. The least time is $\dfrac{5.44 \times 10^{7}}{2.01 \times 10^{8}\ \text{s}^{-1}} = 0.27\ \text{s}$ .

Step 6. The larger sphere (g). It reaches the same potential, $1.57\ \text{V}$ , but its capacitance is twice as large, so $Q = 17.4\ \text{pC}$ .

Answer: (a) $6.22\ \text{eV}$ . (b) $1.57\ \text{eV}$ . (c) $1.57\ \text{V}$ . (d) $8.70\ \text{pC}$ . (e) $5.44 \times 10^{7}$ . (f) $0.27\ \text{s}$ . (g) $17.4\ \text{pC}$ .

Check. The student is wrong. More light, or a larger sphere, frees more electrons per second, and so charges the sphere sooner, but the final potential is fixed by the fastest electron alone: it is the stopping potential, $1.57\ \text{V}$ .

Watch out. The final potential is not the photon energy, $6.22\ \text{V}$ , nor the work function: the work function is spent at the surface, and only the energy left over can carry an electron away against the charge.

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