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. $A{}^{238}U$nucleus, which decays by alpha emission, is 15fm in diameter. Model an alpha particle within a ${}^{238}U$nucleus as being in a one-dimensional box. What is the maximum speed 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 ${}^{238}U$nucleus?

Short answer

a.) The maximum speed is $\mathbf{1}\mathbf{.}\mathbf{66}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}$m/s.

b.) The frequency (reflections/s) for a maximum-speed alpha particle within a ${}^{238}U$nucleus is $\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{20}}$Hz

Step-by-step solution

a.) Given Information U238nucleus, which decays by alpha emission, is 15fm in diameter. Need to find maximum speed an alpha particle  is likely to have

a.) The uncertainty in knowing the position of the alpha particle inside a one-dimensional box of a length of ( 15 fm ) is ( $∆$x = 15 fm ).The uncertainty in the velocity can be calculated using the Heisenberg uncertainty principle

localid="1650900175806" $\mathbf{∆}\mathit{x}\mathbf{∆}{\mathit{p}}_{\mathbf{x}\mathbf{}}\mathbf{=}\frac{\mathbf{h}}{\mathbf{2}}$

Assume the most accurate measurements so $\left(\ge \right)$can be replaced with (=). Moreover, we know that $∆p_x=m∆v_x$and the mass of the $\alpha$particle is 6.644 $\times {10}^{-27}$kg. Thus

$\mathbf{∆}{\mathit{v}}_{\mathbf{x}}\mathbf{=}\frac{\mathbf{h}}{\mathbf{2}\mathbf{m}\mathbf{∆}\mathbf{x}}\mathbf{=}\frac{\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{34}}\mathbf{}\mathbf{J}\mathbf{.}\mathbf{s}}{\mathbf{2}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{644}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{}\mathbf{k}\mathbf{g}\mathbf{)}\mathbf{}\mathbf{(}\mathbf{15}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{m}\mathbf{)}}$=role="math" localid="1650902778537" $\mathbf{3}\mathbf{.}\mathbf{32}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$

which means that the range of possible velocities for the electron is fromrole="math" localid="1650903455497" $(-v_{\mathrm{x}}/2=-1.66\times {10}^{6}m/s)$torole="math" localid="1650903473898" $(∆v_{\mathrm{x}}/2=1.66\times {10}^{6}m/s)$, so the maximum speed that an alpha particle is likely to have is localid="1650903499197" role="math" $1.66\times {10}^{6}m/s.$

b.) Given Information :  frequency (reflections/s) for a maximum-speed alpha particle within a nucleus

b.) Since we are dealing with a one-dimensional model, then dividing the maximum speed of the alpha particle by the length of the box, namely the diameter of the nucleus, will give us the number of reflections of the alpha particle in one second, which is the frequency

$f=\frac{v_x}{L}=\frac{1.66\times {10}^{6}m/s}{15\times {10}^{-15}m}=1.1\times {10}^{20}Hz$