As a scientist in a nuclear physics research lab, you are conducting a photodisintegration experiment to verify the binding energy of a deuteron. A photon with wavelength l in the air is absorbed by a deuteron, which breaks apart into a neutron and a proton. The two fragments share the released kinetic energy equally, and the deuteron can be assumed to be initially at rest.

You measure the speed of the proton after the disintegration as a function of the wavelength l of the photon. Your experimental results are given in the table.

L 110_13 m2 3.50 3.75 4.00 4.25 4.50 4.75 5.00

v1106 m,s 2 11.3 10.2 9.1 8.1 7.2 6.1 4.9

(a) Graph the data as y2 versus 1>l. Explain why the data points, when graphed this way, should follow close to a straight line. Find the slope andy-intercept of the straight line that gives the best fit

to the data.

(b) Assume that h and c have their accepted values. Use your results from part (a) to calculate the mass of the proton and the binding energy (in MeV) of the deuteron.

Let us relist the given data in the table and add two more rows to list the values of $v^2$and$1/\lambda$

Now, we graph the data as $v^2$versus $1/\lambda$and obtain the best fit line as shown below.

When the photon is absorbed by the deuteron, part of its energy works to

separate the proton from the neutron;

This energy equals the binding energy of the deuteron

The rest of the energy is carried by the proton and the neutron as kinetic energy.

the energy of a photon of wavelength$\lambda$ is:

$E_{\gamma }=\frac{hc}{\lambda }$

If the kinetic energy of the proton is$K_p$and that of the neutron is$K_n$, thus

$\frac{hc}{\lambda }=E_b+K_P+K_n$

But we are given that the kinetic energy is shared equally between the proton and the neutron,

$K_p=K_n$

So

$\frac{hc}{\lambda }=E_b+2K_p$

$\frac{hc}{\lambda }=E_b+m_p{v}^2$

Where v is the speed of the proton. Rearranging this equation, we get:

$v^2=\frac{hc}{m_p}\frac{1}{\lambda }-\frac{E_b}{m_p}$

, , h and c are constants, and only v and are variables.

So, graphing $v^2$verses $1/\lambda$should give us a straight line since $v^2$is directly proportional to$1/\lambda$

From the graph, the slope of the line is slope =$119.88m^3/s^2$ and the y-intercept of the straight line is$y_o=-2.1562\times {10}^{14}{m}^3/s^2$

From equation (i), the slope of the line is:

$$\begin{gathered} slope=\frac{hc}{m_p} \\ {m}_p=\frac{hc}{slope} \end{gathered}$$

So, we substitute our values for h =$6.626\times {10}^{-34}\text{J}·\text{s}$ , c = $3.0\times {10}^8\text{m}/\text{s}$, and slope, so we get:

$$\begin{gathered} m_p=\frac{(6.626\times {10}^{-34}J/s)(3.0\times {10}^{8}m/s)}{119.88m^3/s^2} \\ =1.66\times {10}^{-27}Kg \end{gathered}$$

Also, from equation (i), the y-intercept of the straight line is:

$$\begin{gathered} y_o=\frac{E_b}{m_p} \\ {E}_b=-m_p{y}_o \end{gathered}$$

So, we substitute our values for $m_p$and $y_0$, so we get the binding energy of deuteron:

$$\begin{gathered} E_b=(1.66\times {10}^{-27}Kg)(-2.156\times {10}^{14}{m}^2/s^2) \\ =3.58\times {10}^{-13}J \\ {E}_b=(3.58\times {10}^{-13}J)\left(\frac{1.00MeV}{1.60\times {10}^{-13}J}\right) \\ =2.235MeV \end{gathered}$$