(i) transmutation as it relates to radioactivity; (ii) stopping potential.
(c) A certain metal of work function 1.6 eV is irradiated with ultra-violet light of wavelength 3.6 x 10\(^{-7}\) Calculate the maximum
(ii) speed of an emitted electron. (1eV = 1.6 x 10\(^{-18}\) J; C = 3.0 x 10\(^{8}\) ms\(^{-1}\); m, = 9.1 x 10\(^{-31}\) kg; h = 6.6 x 10\(^{-34}\) Js)
(d) If source of the ultra-violet light in (c) above is mo away from the surface of the metal, state the of on the maximum speed of the ejected electron
(a)(i) Transmutation: Transmutation is the change of one element into another as a result of a nuclear reaction, in which the number of protons in the nucleus (the atomic number) changes, for example during radioactive decay or artificial nuclear bombardment.
(a)(ii) Stopping potential: The stopping potential is the minimum negative (retarding) potential that must be applied to the collecting electrode of a photocell to just stop the most energetic photoelectrons from reaching it, so that the photocurrent becomes zero. \(eV_s = KE_{max}\).
(b) Nuclear equations:
\[ {}^{23}_{11}A + {}^{2}_{1}B \rightarrow {}^{p}_{q}C + {}^{1}_{1}\text{(proton)}. \]
Conserving nucleon number: \(23 + 2 = p + 1 \Rightarrow p = 24\). Conserving charge: \(11 + 1 = q + 1 \Rightarrow q = 11\).
\[ {}^{24}_{11}C \rightarrow {}^{r}_{s}E + {}^{0}_{-1}\beta. \]
Conserving nucleon number: \(24 = r + 0 \Rightarrow r = 24\). Conserving charge: \(11 = s + (-1) \Rightarrow s = 12\).
Therefore \(p = 24,\; q = 11,\; r = 24,\; s = 12\).
(c) Energy of the incident photon:
\[ E = \frac{hc}{\lambda} = \frac{(6.6\times10^{-34})(3.0\times10^{8})}{3.6\times10^{-7}} = 5.5\times10^{-19}\,\text{J}. \]
Work function \(W_0 = 1.6\,\text{eV} = 1.6\times(1.6\times10^{-19}) = 2.56\times10^{-19}\,\text{J}\).
(i) Maximum kinetic energy:
\[ KE_{max} = E - W_0 = 5.5\times10^{-19} - 2.56\times10^{-19} = 2.94\times10^{-19}\,\text{J}. \]
(ii) Speed of the emitted electron: from \(KE = \tfrac{1}{2}mv^{2}\),
\[ v = \sqrt{\frac{2\,KE}{m}} = \sqrt{\frac{2\times2.94\times10^{-19}}{9.1\times10^{-31}}} = \sqrt{6.46\times10^{11}} \approx 8.04\times10^{5}\,\text{m s}^{-1}. \]
(d) Moving the source of ultra-violet light farther from the metal only reduces the intensity (number of photons per second arriving), so fewer electrons are emitted per second. It does not change the frequency/energy of each photon, so the maximum kinetic energy and hence the maximum speed of the ejected electron remain unchanged.
