Answer with Explanation:
We are given that
Potential energy=10 MeV
Energy of proton=E=2 MeV
Thickness of energy barrier=L=8.4fm=[tex]8.4\times 10^{-15} m[/tex]
[tex]1 fm=10^{-15} m[/tex]
a.[tex]b=\sqrt{\frac{8\pi^2m(U-E)}{h^2}}[/tex]
[tex]b=\sqrt{\frac{8\pi^2\times 1.67\times 10^{-27}\times (10-2)\times 1.6\times 10^{-19}}{(6.626\times 10^{-34})^2}[/tex]
Where 1 MeV=[tex]1.6\times 10^{-19}\times 10^6 V[/tex]
[tex]h=6.626\times 10^{-34}[/tex]
Mass of proton=[tex]m=1.67\times 10^{-27} kg[/tex]
[tex]b=6.2\times 10^{14}/m[/tex]
T=[tex]e^{-2bL}=e^{-2\times 6.2\times 10^{14}\times 8.4\times 10^{-15}}=29.9\times 10^{-6}[/tex]
b.On other side
Potential energy=U=0
Kinetic energy=K.E=2 Me V
[tex]K_T=2 MeV[/tex]
c.Energy of reflected proton=[tex]K_R=K.E=2 MeV[/tex]
d.Mass of deuteron=[tex]2.0141\times 1.66\times 10^{-27} kg[/tex]
[tex]b=\sqrt{\frac{8\pi^2\times 2.0141\times 1.66\times 10^{-27}\times 8\times 1.6\times 10^{-19}\times 10^6}{(6.626\times 10^{-34})^2}[/tex]
[tex]b=8.77\times 10^{14}/m[/tex]
[tex]T=e^{-2\times 8.77\times 10^{14}\times 8.4\times 10^{-15}}[/tex]
[tex]T=3.99\times 10^{-7}[/tex]
By law of conservation of mechanical energy
Kinetic energy of deuteron,[tex]K_t=2 MeV[/tex]
Kinetic energy of reflected deuteron
[tex]K_R=2 MeV[/tex]
