Question

A 2.0 MeV proton is incident on a potential energy barrier of thickness 8.4 fm and height 10 MeV. What are (a) the transmission coefficient T, (b) the kinetic energy Kt the proton will have on the other side of the barrier if it tunnels through the barrier, and (c) the kinetic energy Kr

it will have if it reflects from the barrier? A 2.0 MeV deuteron (the same charge but twice the mass as a proton) is incident on the same barrier. What are (d) T, (e)
Kt
, and (f)
Kr
?

Answer 1

Answer with Explanation:

We are given that

Potential energy=10 MeV

Energy of proton=E=2 MeV

Thickness of energy barrier=L=8.4fm=
`8.4* 10^(-15) m`

`1 fm=10^(-15) m`

a.
`b=\sqrt{(8\pi^2m(U-E))/(h^2)}`

`b=\sqrt{(8\pi^2* 1.67* 10^(-27)* (10-2)* 1.6* 10^(-19))/((6.626* 10^(-34))^2)`

Where 1 MeV=
`1.6* 10^(-19)* 10^6 V`

`h=6.626* 10^(-34)`

Mass of proton=
`m=1.67* 10^(-27) kg`

`b=6.2* 10^(14)/m`

T=
`e^(-2bL)=e^{-2* 6.2* 10^(14)* 8.4* 10^(-15)}=29.9* 10^(-6)`

b.On other side

Potential energy=U=0

Kinetic energy=K.E=2 Me V

`K_T=2 MeV`

c.Energy of reflected proton=
`K_R=K.E=2 MeV`

d.Mass of deuteron=
`2.0141* 1.66* 10^(-27) kg`

`b=\sqrt{(8\pi^2* 2.0141* 1.66* 10^(-27)* 8* 1.6* 10^(-19)* 10^6)/((6.626* 10^(-34))^2)`

`b=8.77* 10^(14)/m`

`T=e^{-2* 8.77* 10^(14)* 8.4* 10^(-15)}`

`T=3.99* 10^(-7)`

By law of conservation of mechanical energy

Kinetic energy of deuteron,
`K_t=2 MeV`

Kinetic energy of reflected deuteron

`K_R=2 MeV`