Chapter 7, Problem 5P

(a)

To determine

Show that the probability of transmission is $P\approx 16\frac{E}{U}{e}^{-2\left[\sqrt{2m\left(U-E\right)}/\sout{h}\right]L}$ .

Answer to Problem 5P

The probability of transmission $P\approx 16\frac{E}{U}{e}^{-2\left[\sqrt{2m\left(U-E\right)}/\sout{h}\right]L}$ is proved.

Explanation of Solution

Write the expression for transmission probabality.

`    $\frac{1}{T\left(E\right)}=1+\left[\frac{U^2}{4E\left(U-E\right)}\right]\sin h^2\alpha L$        (I)

Here, $T\left(E\right)$ is the transmission probabality and $\alpha$ is the wave number.

Write the expression for $\alpha$ .

    $\alpha ={\left[\frac{2m}{{\sout{h}}^2}\left(U-E\right)\right]}^{1/2}$

Here, $m$ is the mass of proton, $E$ is the Kinetic energy and $\sout{h}$ is the Planck’s constant.

For $E<<U$.

${\left(\alpha L\right)}^2=\frac{2mU{L}^2}{{\sout{h}}^2}$

By hypothesis: $\frac{2mU{L}^2}{{\sout{h}}^2}>>1$

Therefore, it can be written that $\sinh \alpha L\approx \frac{1}{2}{e}^{\alpha L}$ and $U-E\approx U$ .

Using above results equation (I) becomes:

    $\begin{array}{c}\frac{1}{T\left(E\right)}\approx 1+\left(\frac{U}{16E}\right)e^{2\alpha L}\\ \approx \left(\frac{U}{16E}\right)e^{2\alpha L}\end{array}$

Therefore probabality is the reciprocal of above expression.

    $\begin{array}{l}P=\left(\frac{16E}{U}\right)e^{-2\alpha L}\\ P\approx 16\frac{E}{U}{e}^{-2\left[\sqrt{2m\left(U-E\right)}/\sout{h}\right]L}\end{array}$

Conclusion:

Thus, the probability of transmission $P\approx 16\frac{E}{U}{e}^{-2\left[\sqrt{2m\left(U-E\right)}/\sout{h}\right]L}$ is proved.

(b)

To determine

The numerical estimates for the exponential factor in P for each of the following cases.

Answer to Problem 5P

The numerical estimates for the exponential factor in P in case 1 is $0.90$, case 2 is $0.36$, case 3 is $0.415$, case 4 is $0$.

Explanation of Solution

Write the expression for probability.

   $P\approx 16\frac{E}{U}{e}^{-2\left[\sqrt{2m\left(U-E\right)}/\sout{h}\right]L}$

Write the exponential factor in P

    $\text{Factor}=e^{-2\left[\sqrt{2m\left(U-E\right)}/\sout{h}\right]L}$        (II)

Case 1:

When $U-E=0.01\text{eV}$ and $L=0.1\text{nm}$.

Substitute $9.11\times {10}^{-31}\text{kg}$ for $m$ , $1.055\times {10}^{-34}\text{Js}$ for $\sout{h}$, $1.60\times {10}^{-21}\text{J}$ for $U-E$ and ${10}^{-10}\text{m}$  for L in equation (II).

$\begin{array}{c}\text{Factor}=e^{-2\left[\sqrt{2\left(9.11\times {10}^{-31}\text{kg}\right)\left(1.60\times {10}^{-21}\text{J}\right)}/1.055\times {10}^{-34}\text{Js}\right]\left({10}^{-10}\text{m}\right)}\\ =e^{-0.10235}\\ =0.90\end{array}$

Case 2:

When $U-E=1\text{eV}$ and $L=0.1\text{nm}$.

Substitute $9.11\times {10}^{-31}\text{kg}$ for $m$ , $1.055\times {10}^{-34}\text{Js}$ for $\sout{h}$, $1.60\times {10}^{-19}\text{J}$ for $U-E$ and ${10}^{-10}\text{m}$  for L in equation (II).

$\begin{array}{c}\text{Factor}=e^{-2\left[\sqrt{2\left(9.11\times {10}^{-31}\text{kg}\right)\left(1.60\times {19}^{-19}\text{J}\right)}/1.055\times {10}^{-34}\text{Js}\right]\left({10}^{-10}\text{m}\right)}\\ =e^{-1.0235}\\ =0.36\end{array}$

Case 3:

When particle is alpha particles whose mass is $m=6.7\times {10}^{-27}\text{kg}$ , $U-E={10}^6\text{eV}$ and $L={10}^{-15}\text{m}$.

Substitute $6.7\times {10}^{-27}\text{kg}$ for $m$ , $1.055\times {10}^{-34}\text{Js}$ for $\sout{h}$, $1.60\times {10}^{-13}\text{J}$ for $U-E$ and ${10}^{-15}\text{m}$  for L in equation (II).

$\begin{array}{c}\text{Factor}=e^{-2\left[\sqrt{2\left(6.7\times {10}^{-27}\text{kg}\right)\left(1.60\times {19}^{-13}\text{J}\right)}/1.055\times {10}^{-34}\text{Js}\right]\left({10}^{-15}\text{m}\right)}\\ =e^{-0.878}\\ =0.415\end{array}$

Case 4:

When object is bowling ball whose mass is $m=8.0\text{kg}$ , $U-E=1\text{J}$ and $L=2\text{cm}$.

Substitute $8.0\text{kg}$ for $m$ , $1.055\times {10}^{-34}\text{Js}$ for $\sout{h}$, $1\text{J}$ for $U-E$ and $0.02\text{m}$  for L in equation (II).

$\begin{array}{c}\text{Factor}=e^{-2\left[\sqrt{2\left(8.0\text{kg}\right)\left(1\text{J}\right)}/1.055\times {10}^{-34}\text{Js}\right]\left(0.02\text{m}\right)}\\ =e^{-1.5\times {10}^{33}}\\ \approx 0\end{array}$

Conclusion:

Thus, the numerical estimates for the exponential factor in P in case 1 is $0.90$, case 2 is $0.36$, case 3 is $0.415$, case 4 is $0$.