Answer :
Answer:
0.08
Explanation:
The alpha particle suffers a head-on collision with the gold nucleus, so it retraces it path after the collision.
Let us take the masses of the particles in atomic mass units.
The initial momentum and kinetic energy of the gold nucleus is 0(since it is stationary). So, applying conservation of momentum and energy, we get the following two equations:
[tex]m_{1}u_{1}=m_{1}v_{1}+m_{2}v_{2}[/tex] ..........(1)
[tex]\frac{1}{2}m_{1}u_{2}^{2}=\frac{1}{2} m_{1} v_{1}^{2} +\frac{1}{2} m_{2} v_{2}^{2}[/tex] ..........(2)
where,
[tex]m_{1}[/tex] = mass of the alpha particle = 4 units
[tex]m_{2}[/tex] = mass of the gold nucleus = 197 units
[tex]u_{1}[/tex] = initial velocity of the alpha particle
[tex]v_{1}[/tex] = final velocity of the alpha particle
[tex]v_{2}[/tex] = final velocity of the gold nucleus
Now, we shall substitute the value of [tex]v_{2}[/tex] from equation (1) in equation (2). After some simplifications, we get,
[tex]u_{1}^{2}=v_{1}^{2}+\frac{m_{1}}{m_{2}} (u_{1}^{2}+v_{1}^{2}-2u_{1}v_{1})[/tex]
Dividing both sides by [tex]u_1^2[/tex] and substituting [tex]x=\frac{v_1}{u_1}[/tex] and [tex]k=\frac{m_1}{m_2}[/tex] , we get,
[tex]1=x^2+k(1+x^2-2x)\\[/tex]
or, [tex]x^2(k+1)-2kx+(k-1)=0[/tex]
Here, [tex]k=\frac{m_1}{m_2}=\frac{4}{197}=0.02[/tex]
Therefore, [tex]x=\frac{2(0.02)\pm\sqrt{(2\times0.02)^2-(4\times1.02\times-0.98)} }{2\times1.02}[/tex]
or, [tex]x = 1, -0.96[/tex]
Our required solution is -0.96 because the final velocity([tex]v_1[/tex]) of the alpha particle will be a little less the initial velocity([tex]u_1[/tex]). The negative sign comes as the alpha particle reverses it's direction after colliding with the gold nucleus.
Fractional change in kinetic energy is given by,
[tex]\delta E=\frac{\frac{1}{2} m_1u_1^2-\frac{1}{2}m_1v_1^2 }{\frac{1}{2}m_1u_1^2 }=1-x^2=0.078\approx0.08[/tex]