Question

Particle accelerators fire protons at target nuclei so that investigators can study the nuclear reactions that occur. In one experiment, the proton needs to have 20 MeV of kinetic energy as it impacts a 207 Pb nucleus. With what initial kinetic energy (in MeV) must the proton be fired toward the lead target? Assume the nucleus stays at rest

Short answer

The initial kinetic energy of proton is$34.2Mev$

Step-by-step solution

Given Information:

Calculate the kinetic energy of the photon by applying law of conservation of energy.

Assume the lead nucleus is at stationary position and proton is moving with $v_1$velocity toward the lead nucleus is at stationary position and proton alpha particle turned back with velocity.

$$\begin{gathered} r_1=r_p+r_{pb}..........\left(1\right) \\ Here,r_{p}isradiusofproton{r}_{pb}isradiusofleadnucleus \end{gathered}$$

The radius of proton can be expressed as

localid="1649752823191" $r_p=\left(1.2fm\right){\left(A_p\right)}^{\frac{1}{3}}$

Here, A is atomic mass number.

Given Expression:

$$\begin{gathered} r_p=\left(1.2fm\right){\left(1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ =1.20fm \end{gathered}$$

The radius of lead nucleus can be expressed as,

$r_p=\left(1.2fm\right){\left(A_p\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

substitute 207 for $A_{pb}$in the equation

$r_{pb}=(1.2fm){\left(207\right)}^{\frac{1}{3}}=7.10fm$

Substitute $1.20fm$for $r_p$and $7.10fm$for $r_{pb}$in equation (1)

$$\begin{gathered} r_j=1.20fm+7.20fm \\ =8.30fm \end{gathered}$$

Convert the unit of distance from $fmtom$

$$\begin{gathered} r_f=8.30fm\left(\frac{{10}^{-15}}{1fm}\right) \\ =8.30x{10}^{-15}m \end{gathered}$$

Given Expression:

$K_1+U_1=K_f+U_{f..........\left(2\right)}$

The potential energy between proton and lead nucleus can be expressed as

$U=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q_1{q}_{ro}}{r_f}$

The kinetic energy of the particle can be expressed as,

$K=\frac{1}{2}m{v}^2$

Now equation (2) changes as

localid="1649753207568" $K_1+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q_1{q}_{ro}}{r_1}=K_1+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q_1{q}_{ro}}{r_1}$

Substitute $r=0$

$$\begin{gathered} K_1+0J=K_{1+}\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q_1{q}_{ro}}{r_1} \\ {K}_{1=K_1+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q_1{q}_{ro}}{r_1}} \\ K=34.2MeV \end{gathered}$$

The initial kinetic energy of proton is $34.2MeV$