A glass bulb holds hydrogen gas. The atoms are excited to the level n = 4 and fall back to lower…

A glass bulb holds hydrogen gas. The atoms are excited to the level $n = 4$ and fall back to lower levels in every possible way, so every spectral line is produced. The gas is thin, so each photon leaves the bulb without being absorbed again. All the light from the bulb falls on a caesium plate of work function $2.14\ \text{eV}$ in an evacuated tube, as in the figure. A collector in the tube is joined to the plate through a variable voltage supply and a microammeter. Use $13.6\ \text{eV}$ for the ground state energy of hydrogen, $h = 6.63 \times 10^{-34}\ \text{J s}$ , $e = 1.6 \times 10^{-19}\ \text{C}$ and an electron mass of $9.1 \times 10^{-31}\ \text{kg}$ .

A schematic of an experiment. On the left, a glass bulb labelled hydrogen gas, excited to the fourth level, sends rays of light to the right. The light passes through a window into an evacuated glass tube and falls on a plate marked caesium, work function 2.14 eV. Facing it is a collector plate. The two plates are joined by wires to a variable voltage source and a microammeter, drawn below the tube.

How many different spectral lines can the bulb emit?
How many of those lines can eject photoelectrons from the plate?
What is the greatest kinetic energy of the photoelectrons, in electron volts?
What stopping potential does the line from $n = 4$ to $n = 2$ alone need, in volts?
What is the shortest de Broglie wavelength of any photoelectron, in nanometres?

First list the energy of every line, and compare each with the work function.

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.

(a) Lines. The pairs of levels among $n = 1, 2, 3, 4$ number $\dfrac{4 \times 3}{2} = 6$ . So there are $6$ lines.

(b) Which lines can eject. The energies, from $E = 13.6\ \text{eV}\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$ , are: $4 \to 1$ : $12.75\ \text{eV}$ ; $3 \to 1$ : $12.09\ \text{eV}$ ; $2 \to 1$ : $10.20\ \text{eV}$ ; $4 \to 2$ : $2.55\ \text{eV}$ ; $3 \to 2$ : $1.89\ \text{eV}$ ; $4 \to 3$ : $0.66\ \text{eV}$ . The work function is $2.14\ \text{eV}$ . The first four are above it, and the last two are below it. So $4$ lines eject electrons.

(c) Greatest kinetic energy. $K_{\max} = 12.75\ \text{eV} - 2.14\ \text{eV} = 10.61\ \text{eV}$ .

(d) Stopping potential. For the $4 \to 2$ line, $eV_0 = 2.55\ \text{eV} - 2.14\ \text{eV} = 0.41\ \text{eV}$ , so $V_0 = 0.41\ \text{V}$ .

(e) De Broglie wavelength. $K = 10.61\ \text{eV} \times 1.6 \times 10^{-19}\ \text{J eV}^{-1} = 1.70 \times 10^{-18}\ \text{J}$ . $p = \sqrt{2mK} = 1.758 \times 10^{-24}\ \text{kg m s}^{-1}$ , so $\lambda = \dfrac{6.63 \times 10^{-34}\ \text{J s}}{1.758 \times 10^{-24}\ \text{kg m s}^{-1}} = 0.377\ \text{nm}$ .

Answer: (a) $6$ . (b) $4$ . (c) $10.61\ \text{eV}$ . (d) $0.41\ \text{V}$ . (e) $0.377\ \text{nm}$ .

Check. The shortcut $\lambda = \dfrac{1.227\ \text{nm}}{\sqrt{K/\text{eV}}}$ for an electron gives $\dfrac{1.227}{\sqrt{10.61}}\ \text{nm} = 0.377\ \text{nm}$ , the same as (e) to three decimals.

Watch out. All $6$ lines are emitted, but the two with energies of $1.89\ \text{eV}$ and $0.66\ \text{eV}$ are below the work function. A line from $n = 3$ to $n = 2$ is the red line of hydrogen, and it cannot eject an electron from caesium.

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