Question

Heavy nuclei often undergo alpha decay in which they emit an alpha particle (i.e., a helium nucleus). Alpha particles are so tightly bound together that it's reasonable to think of an alpha particle as a single unit within the nucleus from which it is emitted.

a. ${\mathrm{A}}^{238}\mathrm{U}$nucleus, which decays by alpha emission, is $15\mathrm{fm}$in diameter. Model an alpha particle within ${}^{238}\mathrm{U}$nucleus as being in a onc-dimensional box. What is the maximum specd an alpha particle is likely to have?

b. The probability that a nucleus will undergo alpha decay is proportional to the frequency with which the alpha particle reflects from the walls of the nucleus. What is that frequency (reflections/s) for a maximum-speed alpha particle within a ${}^{238}\mathrm{U}$nucleus?

Short answer

(a) Thus, the maximum speed of an alpha particle is from $0\text{to}1.7\times {10}^6\mathrm{m}/\mathrm{s}\text{.}$

(b) The frequency for a maximum speed of an alpha particle within the nucleus is $1.13\times {10}^{20}\mathrm{reflection}/\mathrm{s}$

Step-by-step solution

Part (a) Step 1: Given Information

Diameter of an alpha emission $L=15\mathrm{fm}$

$$\begin{gathered} =15\mathrm{fm}\left(\frac{{10}^{-13}\mathrm{m}}{1\mathrm{m}}\right) \\ \mathrm{L}=15\times {10}^{-15}\mathrm{m} \end{gathered}$$

Part (a) Step 2: solution

(a) The velocity is calculated from eq(1)

Applying $\mathrm{h}=6.626\times {10}^{-34}\mathrm{J},\mathrm{m}=4\left(1.670\times {10}^{-27}\mathrm{kg}\right)$and $\mathrm{L}=15\times {10}^{-15}\mathrm{m}$

$$\begin{gathered} \Delta vx=\frac{6.626\times {10}^{-3}J}{2\left(4\left(1.670\times {10}^{-21}\right)\right)\left(\left(15\times {10}^{-13}\mathrm{m}\right)\right)} \\ \Delta vx=3.31\times {10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the velocity within the interval is $-1.7\times {10}^6\mathrm{m}/\mathrm{s}\le \mathrm{vx}\le 1.7\times {10}^6\mathrm{m}/\mathrm{s}$

The velocity is $1.7\times {10}^6\mathrm{m}/\mathrm{s}$

Conclusion:Thus, the maximum speed of an alpha particle is from 0 to $1.7\times {10}^6\mathrm{m}/\mathrm{s}$.

Part (b) Step 1: Given Information

The frequency for maximum speed of an alpha particle within the nucleus is calculated from the formula,

$f=\frac{vx}{L}$

Part (b) Step 2: solution

Applying vx $=1.7\times {10}^6\mathrm{m}/\mathrm{s}$and $L=15\times {10}^{-15}\mathrm{m}$

$\mathrm{f}=\frac{1.7\times {10}^{6\mathrm{m}/\mathrm{s}}}{15\times {10}^{-15}\mathrm{m}}$

$\mathrm{f}=1.{13}^{*}{10}^{20}\mathrm{reflection}/\mathrm{s}$

Conclusion:The frequency for a maximum speed of an alpha particle within the nucleus is$1.13\times {10}^{20}\mathrm{reflection}/\mathrm{s}.$