A nuclide X, with Z = 88 and mass number 226, is at rest and decays by emitting an alpha particle…

A nuclide X, with $Z = 88$ and mass number $226$ , is at rest and decays by emitting an alpha particle. The decay scheme in the figure shows what happens to the daughter nuclide Y, with $Z = 86$ and mass number $222$ . In $94\%$ of the decays Y is left in its ground level. In the other $6\%$ , Y is left in an excited level $0.186\ \text{MeV}$ above the ground level and then goes to the ground level by emitting a gamma ray. The atomic masses are $226.025410\ \text{u}$ for X, $222.017578\ \text{u}$ for Y and $4.002603\ \text{u}$ for helium-4. Use $1\ \text{u} = 931.5\ \text{MeV}/c^2 = 1.66 \times 10^{-27}\ \text{kg}$ , $1\ \text{MeV} = 1.6 \times 10^{-13}\ \text{J}$ and $hc = 1240\ \text{MeV fm}$ . Treat masses of the particles as their mass numbers in $\text{u}$ when you share out the energy. The nucleus X is free and at rest before the decay.

A decay scheme with energy upwards, not to scale. A parent nuclide X, with 88 protons and mass number 226, stands at the top. Below it are two levels of the daughter nuclide Y with 86 protons and mass number 222: the ground level, and an excited level marked 0.186 MeV above it. A long arrow labelled alpha one, 94 per cent, goes from X to the ground level. A shorter arrow labelled alpha two, 6 per cent, goes from X to the excited level. A wavy arrow labelled gamma goes from the excited level down to the ground level.

How much energy is released when Y is left in its ground level?
What is the kinetic energy of the alpha particle in that decay?
What is the kinetic energy of the alpha particle when Y is left in the excited level?
What is the wavelength of the gamma ray?
What is the recoil speed of Y in the decay to the ground level?

The worked solution

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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) Energy released. The mass defect is $226.025410\ \text{u} - 222.017578\ \text{u} - 4.002603\ \text{u} = 0.005229\ \text{u}$ . So $Q = 0.005229\ \text{u} \times 931.5\ \text{MeV}/\text{u} = 4.87\ \text{MeV}$ .

(b) Alpha particle. Equal momenta mean $\dfrac{K_\alpha}{K_Y} = \dfrac{m_Y}{m_\alpha} = \dfrac{222}{4}$ . So the alpha particle takes $\dfrac{222}{226}$ of $Q$ : $K_\alpha = 4.87\ \text{MeV} \times \dfrac{222}{226} = 4.78\ \text{MeV}$ , and Y keeps $0.086\ \text{MeV}$ .

(c) The other branch. The energy left for motion is $4.87\ \text{MeV} - 0.186\ \text{MeV} = 4.68\ \text{MeV}$ . The alpha particle takes $\dfrac{222}{226}$ of it: $4.60\ \text{MeV}$ .

(d) Gamma ray. $\lambda = \dfrac{hc}{E} = \dfrac{1240\ \text{MeV fm}}{0.186\ \text{MeV}} = 6667\ \text{fm} = 6.67\ \text{pm}$ .

(e) Recoil speed. $K_Y = 0.086\ \text{MeV} = 1.38 \times 10^{-14}\ \text{J}$ , and $m_Y = 222 \times 1.66 \times 10^{-27}\ \text{kg} = 3.69 \times 10^{-25}\ \text{kg}$ . So $v = \sqrt{\dfrac{2K_Y}{m_Y}} = 274\ \text{km s}^{-1}$ .

Answer: (a) $4.87\ \text{MeV}$ . (b) $4.78\ \text{MeV}$ . (c) $4.60\ \text{MeV}$ . (d) $6.67\ \text{pm}$ . (e) $274\ \text{km s}^{-1}$ .

Check. By momentum: The alpha particle has mass $4 \times 1.66 \times 10^{-27}\ \text{kg}$ and energy $4.78\ \text{MeV}$ , so $p = \sqrt{2m_\alpha K_\alpha} = 1.01 \times 10^{-19}\ \text{kg m s}^{-1}$ . Dividing by $m_Y$ gives $274\ \text{km s}^{-1}$ , the same as (e).

Watch out. The alpha particle does not take all of $Q$ . Y is heavy, but it still takes the share $\dfrac{4}{226}$ , about $1.8\%$ . And in the branch to the excited level the gamma energy $0.186\ \text{MeV}$ is not available to the alpha particle.

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