Answer:
17664000 V
Explanation:
We are given that
Diameter of uranium,d=15fm
Radius,r=[tex]\frac{d}{2}=\frac{15}{2}=7.5\times 10^{-15}m[/tex]
1 fm=[tex]10^{-15} m[/tex]
Z=92
Charge on uranium particle=[tex]q_1=92\times 1.6\times 10^{-19}[/tex] C
Because charge on proton=[tex]1.6\times 10^{-19} C[/tex]
Charge on alpha particle=[tex]q_2=2e=2\times 1.6\times 10^{-19} C[/tex]
We have to find the potential difference would be needed to accelerate an alpha particle.
Speed of alpha particle=0
Therefore, k.E of alpha particle=0
Energy gained by alpha particle=[tex]E=q_2 V[/tex]
[tex]E=0-U=U=\frac{kq_1q_2}{r}[/tex]
[tex]q_2V=\frac{kq_1q_2}{r}[/tex]
[tex]V=\frac{kq_1}{r}[/tex]
Where [tex]K=9\times 10^9[/tex]
Substitute the values
[tex]V=\frac{9\times 10^9\times 92\times 1.6\times 10^{-19}}{7.5\times 10^{-15}}[/tex]
[tex]V=17664000 V[/tex]
The potential difference that is required to accelerate an alpha particle is [tex]V = 2.35 \times 10^{19}\; Volts[/tex].
Given the following data:
Radius = [tex]\frac{Diameter}{2} = \frac{15\;fm}{2} = 7.5 \times 10^{-15}\;m[/tex]
We know that a proton has a charge of [tex]1.6 \times 10^{-19}\;C[/tex]
Coulomb constant, k = [tex]9 \times 10^9[/tex]
First of all, we would determine the charge on both an uranium and alpha particle.
For uranium:
[tex]q_1 = 92 \times 1.6 \times 10^{-19}\\\\q_1 = 1.47 \times 10^{-19}\;C[/tex]
For alpha particle:
[tex]q_2 = 2 \times 1.6 \times 10^{-19}\\\\q_2 = 3.2 \times 10^{-19}\;C[/tex]
Next, we would determine the potential difference that is required to accelerate an alpha particle by using this derived formula:
[tex]E = qV[/tex]
[tex]q_2V = F_q\\\\q_2V = \frac{kq_1q_2}{r^2} \\\\q_2Vr^2=kq_1q_2\\\\V=\frac{kq_1}{r^2}[/tex]
Substituting the given parameters into the formula, we have;
[tex]V = \frac{9 \times 10^9 \times 1.47 \times 10^{-19}}{(7.5 \times 10^{-15})^2} \\\\V = \frac{1.323 \times 10^{-9}}{5.625 \times 10^{-29}}\\\\V = 2.35 \times 10^{19}\; Volts[/tex]
