Question

when a proton of mass mp is fired on a deutron of mass md having E binding energy splits into ints constituents. find the minimum KE of proton required for this reaction. take mass of neutron mn=mp.

Answer 1

T_{\rm min} = \frac{E(6m_p - E)}{2(2m_p - E)}.

Answer 2

We wish to find the minimum kinetic energy (in the lab frame) of a proton (mass $m_p$) incident on a deuteron (mass

$$m_d=2m_p - E$$

since the deuteron is bound by an energy $E$) so that the deuteron “breaks up” into its proton and neutron constituents. (We are given that the neutron mass is $m_n=m_p$; hence the deuteron mass is less than $m_p+m_n$ by the binding energy $E$.)

Let the reaction be written as

$$p + d\to p+p+n.$$

Because the breakup requires that the final state has three nucleons (each with rest mass $m_p$), the total rest‐mass energy of the products is

$$3m_p.$$

For a reaction at threshold the final products (in the center‐of‐mass system) have no extra kinetic energy; they all move together with the common velocity of the CM. (Even though a three–body final state is more complicated than a two–body case, the threshold condition is found by “putting” the extra available energy to zero in the CM frame apart from that needed to “pay” the binding energy.)

In the lab the deuteron is at rest and the proton has kinetic energy $T$. Thus the total energy (in our units, using $c=1$) is:

$$E_{\rm lab}=m_p+T+(2m_p-E)=3m_p -E+T.$$

The square of the total 4–momentum (the invariant $s$) is given by

$$s = m_p^2 + (2m_p-E)^2 + 2(2m_p -E)(m_p+T).$$

At threshold, the center–of–mass energy equals the total rest–mass of the final products:

$$\sqrt{s}=3m_p \quad \Longrightarrow \quad s=9m_p^2.$$

Thus, we have

$$9m_p^2=m_p^2+(2m_p-E)^2+2(2m_p-E)(m_p+T_{\rm min}).$$

Let’s now solve step‐by–step.

1. Expand $(2m_p-E)^2$:

$$(2m_p-E)^2=4m_p^2-4m_pE+E^2.$$

2. Substitute in the equation:

$$9m_p^2 = m_p^2+4m_p^2-4m_pE+E^2+2(2m_p-E)(m_p+T_{\rm min}).$$

Combine the first three terms:

$$m_p^2+4m_p^2 =5m_p^2,\quad\text{so:}\quad9m_p^2 = 5m_p^2-4m_pE+E^2+2(2m_p-E)(m_p+T_{\rm min}).$$

3. Rearranging gives:

$$9m_p^2-5m_p^2+4m_pE-E^2=2(2m_p-E)(m_p+T_{\rm min}).$$

That is,

$$4m_p^2+4m_pE-E^2=2(2m_p-E)(m_p+T_{\rm min}).$$

4. Solve for $m_p+T_{\rm min}$:

$$m_p+T_{\rm min}=\frac{4m_p^2+4m_pE-E^2}{2(2m_p-E)}.$$

5. Now subtract $m_p$ from both sides:

$$T_{\rm min}=\frac{4m_p^2+4m_pE-E^2}{2(2m_p-E)}-m_p.$$

6. Write $m_p$ with the same denominator:

$$T_{\rm min}=\frac{4m_p^2+4m_pE-E^2-2m_p(2m_p-E)}{2(2m_p-E)}.$$

7. Compute the subtracted term:

$$2m_p(2m_p-E)=4m_p^2-2m_pE.$$

So the numerator becomes:

$$4m_p^2+4m_pE-E^2-4m_p^2+2m_pE=6m_pE-E^2.$$

8. Therefore, the minimum kinetic energy is:

$$T_{\rm min} = \frac{6m_pE-E^2}{2(2m_p-E)} = \frac{E(6m_p-E)}{2(2m_p-E)}.$$

Thus, the required answer is:

$$\boxed{T_{\rm min} = \frac{E(6m_p - E)}{2(2m_p - E)}.}$$

For the typical case where $E \ll 2m_p$ (i.e. the binding energy is small compared to the nucleon mass) one may approximate this as:

$$T_{\rm min}\approx \frac{6m_pE}{4m_p}=\frac{3}{2}E.$$

Below is a mermaid flowchart summarizing the key steps:

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Summary of the Solution

• Core Steps:
  Write the invariant $s$ for the reaction using energy–momentum conservation, set $\sqrt{s}=3m_p$ (final state rest mass), substitute $m_d=2m_p-E$, solve the resulting equation for $T_{\rm min}$ and simplify.

• Answer:

  $$T_{\rm min} = \frac{E(6m_p - E)}{2(2m_p - E)}.$$

  (For $E\ll 2m_p$, $T_{\rm min}\approx \frac{3}{2}E$.)