Question:A $\mathbf{0}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{MeV}$proton is incident on a potential energy barrier of thickness $\mathbf{10}\mathbf{}\mathbf{fm}$and height $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{MeV}$.What are (a) the transmission coefficient T , (b) the kinetic energy ${\mathbf{K}}_{\mathbf{t}}$ the proton will have on the other side of the barrier if it tunnels through the barrier, and (c) the kinetic energy ${\mathbf{K}}_{\mathbf{r}}$ it will have if it reflects from the barrier? A $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{MeV}$ deuteron (the same charge but twice the mass as a proton) is incident on the same barrier. What are (d) T , (e) ${\mathbf{K}}_{\mathbf{t}}$, and (f) ${\mathbf{K}}_{\mathbf{r}}$?

The energy of the incident proton is $\mathrm{E}=3.0\mathrm{MeV}$.

The thickness of the energy barrier is $\mathrm{L}=10\mathrm{fm}$.

Height of the potential barrier is $U_b=10.0\mathrm{MeV}$.

The mass of deuteron is twice the mass of the photon.

The energy of the incident deuteron is ${\mathrm{E}}_{\mathrm{D}}=3.0\mathrm{MeV}$.

A particle for example electrons, or proton can reflect from any boundary at which its potential energy changes even when classically it would not reflect.

The mechanical energy is conserved in the transmission process.

The mechanical energy is conserved in the reflection process

The transmission coefficient of a particle can be given as.

$T=e^{-2bL}$

Where, $L$ is the length of the potential barrier and $b$can be given as,

role="math" localid="1663155509945" $b=\sqrt{\frac{8{\pi }^{2}m\left(E-U_b\right)}{h^2}}$

Where the mass of the particle is $m$ and $E$is incident energy and $U_b$is the height of the barrier.

Substituting the values for photon particle,

$\begin{array}{rcl}\mathrm{b}& =& \sqrt{\frac{8\times 3.{14}^2\times 1.6\times {10}^{27}\mathrm{kg}\times \left(10-3\right)\mathrm{MeV}\times \left(1.602\times {10}^{-13}\mathrm{J}/\mathrm{MeV}\right)}{{\left(6.626\times {10}^{-34}\right)}^2}}\\ & =& \sqrt{33.701785\times {10}^{28}{\mathrm{m}}^{-2}}\\ & =& 5.8082\times {10}^{14}{\mathrm{m}}^{-1}\end{array}$

Therefore, the transmission coefficient

$\begin{array}{rcl}\mathrm{T}& =& \exp \left(-2\times 5.8\times {10}^{14}{\mathrm{m}}^{-1}\times 10\times {10}^{-15}\mathrm{m}\right)\\ & =& 9.02\times {10}^{-6}\end{array}$

Therefore, the transmission coefficient in the case of a photon is of is $T=9.02x{10}^{-6}$.

From the conservation of the energy, the Mechanical energy is conserved in the transmission process.

Since when the proton reaches the barrier, it has a kinetic energy of $3.0\mathrm{MeV}$ and potential energy of zero.

After passing the potential barrier, the proton again will have a potential energy of zero, and kinetic energy of $3.0\mathrm{MeV}$.

Hence, the kinetic energy $K_t$ of photon is $3.0\mathrm{MeV}$.

From the conservation of the energy, the Mechanical energy is conserved in the reflection process.

Since when the proton reaches the barrier, it has a kinetic energy of $3.0\mathrm{MeV}$and potential energy of zero.

Therefore, after reflection, the proton will have a potential energy of zero, and kinetic energy of $3.0\mathrm{MeV}$

Therefore, the kinetic energy $K_r$of the photon is $3.0\mathrm{MeV}$.

The transmission coefficient of a particle can be given as

$T=e^{-2bL}$

Where, $L$is the length of the potential barrier and $b$can be given as

$b=\sqrt{\frac{8{\pi }^{2}m\left(E-U_b\right)}{h^2}}$

Where the mass of the particle is $m$and $E$is incident energy and data-custom-editor="chemistry" $U_b$is the height of the barrier.

Substituting the values for deuteron particle.

$\begin{array}{rcl}\mathrm{b}& =& \sqrt{\frac{8\times 3.{14}^2\times 1.6\times {10}^{27}\mathrm{kg}\times \left(10-3\right)\mathrm{MeV}\times \left(1.602\times {10}^{-13}\mathrm{J}/\mathrm{MeV}\right)}{{\left(6.626\times {10}^{-34}\right)}^2}}\\ & =& 8.2143\times {10}^{14}{\mathrm{m}}^{-1}\end{array}$

Therefore, the transmission coefficient

$\begin{array}{rcl}\mathrm{T}& =& \exp \left(-2\times 8.2143\times {10}^{14}{\mathrm{m}}^{-1}\times 10\times {10}^{-15}\mathrm{m}\right)\\ & =& 7.33\times {10}^{-8}\end{array}$

Therefore, the transmission coefficient in the case of the deuteron is $T=7.33\times {10}^{-8}$.

From the conservation of the energy, the Mechanical energy is conserved in the transmission process.

Since when the deuteron reaches the barrier, it has a kinetic energy of $3.0\mathrm{MeV}$.

After passing the potential barrier, the deuteron again will have the kinetic energy of$3.0\mathrm{MeV}$.

Therefore, In the case of deuteron the kinetic energy $3.0\mathrm{MeV}$.

From the conservation of the energy, the Mechanical energy is conserved in the reflection process.

Since when the deuteron reaches the barrier, it has the kinetic energy of $3.0\mathrm{MeV}$

Therefore, after reflection, the deuteron will have the kinetic energy of $3.0\mathrm{MeV}$.

Therefore, In the case of deuteron the kinetic energy .