(a) State three conclusions that can be drawn from Rutherford's experiment on the scattering of alpha particles by a thin metal foil in relation to the structure of the atom
The diagram above illustrates th3 energy levels of an electron in an atom. If an excited electron moves from \(n_2\) to \(n_\theta\), calculate the:
(ii) wavelength of the emitted radiation. [ \(h = 6.6 \times 10^{-34}\) Js; le V = .6 \(\times 10^{-19}\) J; C = 3.0 \(\times 10^8\) ms\(^{-1}\)]
(c) The following nuclear equations represent two types of radioactivity.
(a) Three conclusions from Rutherford's alpha-particle scattering experiment
- Most of the atom is empty space, because the vast majority of alpha particles passed straight through the thin metal foil undeflected.
- Each atom has a very small, dense core (the nucleus) which carries the whole positive charge, because a few alpha particles were deflected through large angles and a very small number were bounced almost straight back.
- Nearly all the mass of the atom is concentrated in this tiny nucleus, while the electrons occupy the large space surrounding it.
(b) Transition from \(n_2\) to \(n_0\)
From the energy-level diagram: \(n_2 = -2.0\ \text{eV}\) and \(n_0 = -12.0\ \text{eV}\). The energy of the emitted photon is the difference between these levels:
\[ E = E_{n_2} - E_{n_0} = (-2.0) - (-12.0) = 10.0\ \text{eV} \]
Convert to joules using \(1\ \text{eV} = 1.6\times10^{-19}\ \text{J}\):
\[ E = 10.0 \times 1.6\times10^{-19} = 1.6\times10^{-18}\ \text{J} \]
(i) Frequency from \(E = hf\):
\[ f = \frac{E}{h} = \frac{1.6\times10^{-18}}{6.6\times10^{-34}} = 2.42\times10^{15}\ \text{Hz} \]
(ii) Wavelength from \(c = f\lambda\):
\[ \lambda = \frac{c}{f} = \frac{3.0\times10^{8}}{2.42\times10^{15}} = 1.24\times10^{-7}\ \text{m} \]
(about 124 nm, in the ultraviolet region.)
(c) Types of radioactivity
Equation A: \(^{226}_{88}\text{Ra} \to\ ^{222}_{86}\text{Rn} + ^{4}_{2}\alpha\). This is natural (spontaneous) radioactivity, specifically alpha decay: an unstable nucleus disintegrates on its own, emitting an alpha particle and forming a new element.
Equation B: \(^{14}_{7}\text{N} + ^{4}_{2}\alpha \to\ ^{17}_{8}\text{O} + ^{1}_{1}p\). This is artificial (induced) transmutation: a stable nucleus is deliberately bombarded by an incoming particle (an alpha particle), changing it into a different nucleus.
Difference: In A the disintegration is spontaneous, happening by itself with no external cause; in B the nuclear change is artificially induced by bombarding the target nucleus with a fast-moving particle.
