In a cloud chamber an alpha particle collides elastically…
In a cloud chamber an alpha particle collides elastically with a nucleus of an unknown gas that is practically at rest. After the collision the alpha particle is deflected by $\theta_1 = 8^\circ 27'$ from its original direction, and the unknown particle moves off at $\theta_2 = 60^\circ$ to that direction. Find the mass ratio and identify the particle.
Explanation of Method
Identifying a particle from collision angles.
Step 1: Mass ratio from the angles
Momentum conservation along and across the original direction, together with energy conservation, gives
\[ \frac{m_\alpha}{m_x} = \frac{\sin(\theta_1 + 2\theta_2)}{\sin\theta_1}. \]
Step 2: Numbers
\[ \frac{m_\alpha}{m_x} = \frac{\sin 128^\circ 27'}{\sin 8^\circ 27'} = \frac{0.783}{0.147}. \]
\[\frac{m_\alpha}{m_x} \approx 5.3\]
Step 3: Which particle?
No nucleus has $m_x \approx m_\alpha/5.3 \approx 0.75\ \mathrm{u}$; the lightest is hydrogen (ratio 4). The given angles therefore do not fit exactly. For hydrogen (ratio 4), a recoil at 60° would go with an alpha deflection of 10.9°, and an alpha deflection of 8°27' with a recoil at 67.8°. The data point to the lightest possible target, hydrogen (a proton), but the ratio they give is 5.3, not 4.
Result
\[\boxed{\,\frac{m_\alpha}{m_x} \approx 5.3\,}\]
Remark
About the original solution. Its formula is correct, but $\sin 128^\circ 27' / \sin 8^\circ 27'$ is about 5.3, not 4. The identification as hydrogen is plausible (it is the lightest nucleus and the only candidate below helium), but it does not follow from the numbers as given; one of the angles in the source is probably misprinted. (For comparison, a deflection of 18°27' would give a ratio of 2.1, i.e. deuterium.)