Worked Examples 3

Example 1 (Gamma-decay recoil)

An excited nucleus, initially at rest, undergoes gamma decay: \[ N^* \rightarrow N+\gamma. \] The total energy released is \(Q\), and the rest mass of the recoiling daughter nucleus is \(M\). Let the photon energy be \(E_\gamma\).

  1. What is the magnitude of the momentum of the recoiling nucleus?
  2. What is the total relativistic energy of the recoiling nucleus?
  3. Show that \[ E_\gamma=\frac{Q(2Mc^2+Q)}{2(Mc^2+Q)}. \]
  4. Taking note that \(Q\ll Mc^2\) for nuclear gamma decay, show that \[ E_\gamma \approx Q-\frac{Q^2}{2Mc^2}. \]

Solution 1.

Conservation of momentum requires: \[ \left |\v p_\gamma \right | = \left|\v p_D\right| \]

For the photon, \[ E_\gamma=p_\gamma c = p_D c \] For the recoiling daughter nucleus, \[ E_D^2 =p_D^2c^2+M^2c^4 =E_\gamma^2+M^2c^4. \] Energy conservation gives \[ Q = E_\gamma + \text{KE}_D = E_\gamma + (E_D - Mc^2) \] Therefore, \[ \left(Q-E_\gamma+Mc^2\right)^2 = E_D^2 =E_\gamma^2+M^2c^4. \] Expanding, \[ \left(Q-E_\gamma\right)^2 +2\left(Q-E_\gamma\right)Mc^2 +M^2c^4 =E_\gamma^2+M^2c^4. \] Hence, \[ Q^2-2QE_\gamma+2\left(Q-E_\gamma\right)Mc^2=0, \] so that \[ 2E_\gamma\left(Q+Mc^2\right) =Q^2+2QMc^2. \] Thus, \[ \begin{aligned} E_\gamma &=\frac{Q^2+2QMc^2} {2\left(Q+Mc^2\right)}\\ &=\frac{2Q^2+2QMc^2-Q^2} {2\left(Q+Mc^2\right)} \\ &=\frac{2Q\left(Q+Mc^2\right)} {2\left(Q+Mc^2\right)} -\frac{Q^2}{2\left(Q+Mc^2\right)} \\ &=Q-\frac{Q^2}{2\left(Q+Mc^2\right)}\\ &=Q-\frac{Q^2}{2Mc^2}\frac{1}{\left(Q/Mc^2+1\right)} \\ E_\gamma&\approx Q-\frac{Q^2}{2Mc^2}. \end{aligned} \]

Example 2 (A forbidden decay of the \(\rho^0\))

The \(\rho^0\) meson (neutral, odd intrinsic parity, spin-\(1\)) has several decay channels. However, the following seemingly valid decay of the \(\rho^0\) meson into two \(\pi^0\) mesons (neutral, odd intrinsic parity, spin-\(0\)) is never observed:

\[ \rho^0\rightarrow\pi^0+\pi^0. \]

Explain why this decay is not observed, despite appearing kinematically allowed.

Solution 2 (A forbidden decay of the \(\rho^0\)).

Considering the conservation of angular momentum:

  • Before the decay: The parent \(\rho^0\) is at rest, so there is no external orbital angular momentum. Its total angular momentum is therefore its spin:

\[ J=1. \]

  • After the decay: Both pions have spin zero, so the total angular momentum must come entirely from their relative orbital angular momentum:

    \[ J=\ell=1. \]

  • Hence, in the centre-of-mass frame, the spatial wavefunction of the two-pion system must be proportional to a spherical harmonic \(Y_1^m(\theta,\phi)\).

Considering exchange symmetry

  • The two \(\pi^0\) mesons are identical bosons, so their total wavefunction must be symmetric under the exchange \(\pi^0\longleftrightarrow\pi^0\).

  • Since both pions have spin zero and no other internal quantum number distinguishes them, their two-particle state must have a symmetric spatial wavefunction.

  • Exchanging the two pions reverses their relative position vector:

    \[ \vec{r}\rightarrow-\vec{r}, \]

    corresponding to

    \[ (\theta,\phi)\rightarrow(\pi-\theta,\phi+\pi). \]

  • Under this transformation,

    \[ Y_\ell^m(\pi-\theta,\phi+\pi) =(-1)^\ell Y_\ell^m(\theta,\phi). \]

Therefore, exchange symmetry requires \(\ell\) to be even.

Angular-momentum conservation requires \(\ell=1\), whereas the exchange symmetry requires \(\ell\) to be even. These requirements are incompatible. Thus, although the decay is kinematically allowed, it is forbidden by angular-momentum conservation together with Bose exchange symmetry.

Example 3 (Electron in the deuteron?)

The deuteron is the nucleus of the deuterium atom. It is known to consist of a proton and a neutron. It has charge \(+1\), even parity, and spin \(1\).

An alternative hypothesis suggests that the deuteron is instead composed of two protons and an electron.

Give two independent physical arguments showing that the proton–proton–electron hypothesis is inconsistent with the observed properties of the deuteron.

Solution 3.

Argument 1. Using addition of angular momentum

Let \(\vec{S}_{p_1}\), \(\vec{S}_{p_2}\), and \(\vec{S}_e\) denote the spins of the two protons and the electron. Each constituent has spin \(\frac12\):

\[ s_{p_1}=s_{p_2}=s_e=\frac12. \]

USing the angular-momentum addition rule gives to First combine the spins of the two protons gives:

\[ \vec{J}\,'=\vec{S}_{p_1}+\vec{S}_{p_2}. \]

\[ J'=0,1. \]

Combining this result with the electron spin,

\[ \vec{J}=\vec{J}\,'+\vec{S}_e, \]

gives

\[ J=\frac12 , \frac32 \]

Including orbital angular momentum cannot repair the problem: orbital angular momentum has integer quantum number \(L\), and coupling an integer \(L\) to a half-integer intrinsic spin still produces only half-integer total angular momentum. The proton–proton–electron model therefore cannot reproduce the observed integer spin \(J=1\) of the deuteron.

Argument 2: Using the Uncertainty Principle

To confine an electron to a region of nuclear dimensions \(\Delta x\), the uncertainty principle requires

\[ \Delta p\,\Delta x\gtrsim\frac{\hbar}{2}. \]

I.e.,

\[ (\Delta p)c\gtrsim\frac{\hbar c}{2\Delta x}. \]

The relativistic energy of the confined electron would satisfy

\[ E_e^2=(pc)^2+m_e^2c^4, \]

and hence

\[ E_e \gtrsim \sqrt{ \left(\frac{\hbar c}{2\Delta x}\right)^2 +m_e^2c^4 }. \]

Taking the nuclear dimension \(\Delta x\sim1\ \mathrm{fm}\) and noting that \(\hbar c\simeq197\ \mathrm{MeV\,fm}\) gives

\[ \frac{\hbar c}{2\Delta x} \gtrsim \frac{197\ \mathrm{MeV\,fm}} {2(1\ \mathrm{fm})} \approx 100\ \mathrm{MeV}. \]

Since \(m_ec^2\simeq0.511\ \mathrm{MeV}\), the electron would be highly relativistic, with kinetic energy of order \(100\ \mathrm{MeV}\). This is far greater than the deuteron binding energy of approximately \(2.2\ \mathrm{MeV}\) (see problem set 2).

Therefore, an electron could not be confined inside the deuteron.