Kinetic energy of an alpha particle

[Taylor_1989]

Homework Statement

During the $\alpha$ decay of $^{238}_{92}U$ an $\alpha$ particle is fromed inside the nucleus. Given that the last 4 nucleons in $^{238}_{92}U$ have an average binding energy per nucleon of $6.1MeV$, estimate the kinetic energy of the $\alpha$ particle when it is formed.

Addtional information
Mass of $\alpha=4.00260amu$
Mass of neutron =$1.00866amu$
Mass of Proton = $1.00783amu$
amu = $1.66054\cdot10^{-27}$

Homework Equations

$Q=B.E_f-B.E_i$

3. The Attempt at a Solution
I am been trying visulise this question and the best way I have come with is as follow:

Using the standard alpha decay equation

$$^{232}_92U\rightarrow^{230}_{90}X+\alpha$$

Now if I thing of separating the LHS of the equation I would gee the following

$$B.E_x+mc^2_x +B.E_{\alpha 1}+mc^2_{\alpha 1}=B.E_x+mc^2_x+B.E_{\alpha 2}+mc^2_{\alpha 2}$$

Where $B.E_{\alpha 1}$ is the energy of the last four nucleons.

So from this the LHS $B.E_x+mc^2_x$ will cancel with the $B.E_x+mc^2_x$ on RHS and I will be left with:$$B.E_{\alpha 1}+mc^2_{\alpha 1}=B.E_{\alpha 2}+mc^2_{\alpha 2}$$

So as the total number of protons and neutrons are equal on both sides the the following equation can be formed:

$$mc^2_{\alpha 1}-mc^2_{\alpha 2}=B.E_{\alpha 2}-B.E_{\alpha 1}$$

Where the LHS of the equation is the Q value hence

$$Q=B.E_{\alpha 2}-B.E_{\alpha 1}$$

Calculating the B.E for $B.E_{\alpha 2}$ give the following

$$B.E_{\alpha 2}=2n+2p=28.29MeV$$

pluggin into my Q value equation

$$Q=29.29MeV-4*6.1MeV=3.89MeV$$

Is this the correct method, and through process in solving this equation, or have I viewed it in the wrong way?

 

[kuruman]

The "standard" alpha decay equation $^{232}_{92}U \rightarrow^{230}_{90}X+\alpha$ does not make sense. The number of nucleons on the right side is 230 in $X$ and 4 in the alpha particle for a total of 234. However, you start with 232 on the left side. Furthermore, the question says that you have $^{238}_{92}U$.

You are correct in saying that the $mc^2$ terms will cancel. If I were you, I would calculate the total binding energy on each side and get $Q$ that way. You are given 6.1 MeV/nucleon that you can use for $^{238}_{92}U$ and $^{234}_{90}X$ and you can look up the binding energy of $^{4}_{2}He$. I think you will get the same answer because the daughter nuclide has the same binding energy as the parent.

 

[Taylor_1989]

Ok so I have made some typos in my workings, which I will correct, sorry for that. I will also redo more working in a more clearer manner.

Question:

During the $\alpha$ decay of $^{238}_{92}U$ an $\alpha$ particle is fromed inside the nucleus. Given that the last 4 nucleons in $^{238}_{92}U$ have an average binding energy per nucleon of $6.1MeV$, estimate the kinetic energy of the $\alpha$ particle when it is formed.

My new working are as follows:
$$^{238}_{92}U\rightarrow^{234}_{90}X+\alpha_2$$

My view is that the LHS can be broken down like so:

$$^{234}_{90}X+\alpha_1\rightarrow^{234}_{90}X+\alpha_2 + Q$$

Because I have the binding energy of the last 4 nucleons, which I have designated $alpha_1$, so that leaves the binding energy of the remaining 234 nucleons within that nucleus.

As I am looking for the Q value of the alpha particle I will start with the summation equation for Q

$$Q=\sum _{intial}mc^2-\sum _{final}mc^2$$

I then calculated the mass energy for each individual part for LHS and RHS

$$m(^{234}_{90}X)c^2=90m_p+144m_n)c^2-B.E_x$$
This would be for RHS also

$$m(^4_2\alpha)c^2=2n+2p-(4*6.1MeV)$$

RHS
$$m(^4_2\alpha)c^2=2n+2p-B.E_{\alpha}$$

So subbing into the Q equation that I have given I will have the follwoing

$$Q=B.E_{\alpha}-(4*6.1Mev)$$

Using the calculated value of $B.E_{alpha}=28.1MeV$

I make the Q value $Q=3.7MeV$

I have edited the original question also.