Particles Behaving as Waves Problems and Solutions 9

Q#39.79
(a) A particle with mass m has kinetic energy equal to three times its rest energy. What is the de Broglie wavelength of this particle? (Hint: You must use the relativistic expressions for momentum and kinetic energy: E = $(pc)^2 + (mc^2)^2$  and K = E $-mc^2$) 

(b) Determine the numerical value of the kinetic energy (in MeV) and the wavelength (in meters) if the particle in part (a) is (i) an electron and (ii) a proton. 

Answer:

Combining the two equations in the hint gives

 pc = $\sqrt{K(K+2mc^2)}$ and

$\lambda = \frac{hc}{\sqrt{K(K+mc^2)}}$

(a) With K = 3m$c^2$  this becomes $\lambda = \frac{hc}{\sqrt{(3mc^2)(3mc^2+mc^2)}}=\frac{h}{\sqrt{15}mc}$

(b) (i) K = 3m$c^2 = 2(9.109 \times 10^{-31} \ kg)(2.998 \times 10^8 \ m/s)$ = $2.456 \times 10^{13} J$ = 1.53 MeV

$\lambda=\frac{h}{\sqrt{15}mc}=\frac{6.626 \times 10^{-34} \ J.s}{\sqrt{15}(9.109 \times 10^{-31} \ kg)(2.998 \times 10^8 \ m/s)}$ = $6.26 \times 10^{-13} $ m

(ii) K is proportional to m, so for a proton

K = $\frac{m_p}{m_e}(1.53 \ MeV)$

K = $(1836)(1.53 \ MeV)$ = 2810 MeV

λ is proportional to 1/m so for a proton

$\lambda$ = $\frac{m_p}{m_e}(6.26 \times 10^{-13} \ m)$

$\lambda$ = $\frac{1}{1836}(6.26 \times 10^{-13} \ m)$ = 3.14 $\times 10^{-16}$ m 

#39.80 

Proton Energy in a Nucleus. The radius of atomic nuclei are of the order of 5,0 $\times$ $10^{-15}$ m.

(a) Estimate the minimum uncertainty in the momentum of a proton if it is confined within a nucleus.

(b) Take this uncertainty in momentum to be an estimate of the magnitude of the momentum. Use the relativistic relationship between energy and momentum, Eq. (37.39), to obtain an estimate of the kinetic energy of a proton confined within a nucleus. 

(c) For a proton to remain bound within a nucleus, what must the magnitude of the (negative) potential energy for a proton be within the nucleus? Give your answer in eV and in MeV. Compare to the potential energy for an electron in a hydrogen atom, which has a magnitude of a few tens of eV. (This shows why the interaction that binds the nucleus together is called the “strong nuclear force.”) 

Answer:

Apply the Heisenberg Uncertainty Principle. Consider only one component of position and momentum.

$\Delta x \Delta p_x$ = $\frac{\hbar}{2}$

Take $\Delta x \approx 5.0 \times 10^{-15}$ m. 

K = E $- mc^2$

For a proton, m = 1.67 $\times 10^{-27}$ kg

(a) $\Delta p_x=\frac{\hbar}{2\Delta x}=\frac{1.055 \times 10^{-34} \ J.s}{2(5.0 \times 10^{-15} \ m)}$

$\Delta p_x$ = 1.1 $\times 10^{-20}$ kg.m/s

(b) $mc^2 = (1.67 \times 10^{-27} \ kg)(2.998 \times 10^8 \ m/s)^2 =1.501 \times 10^{-10}$ J  

and pc = $(1.1 \times 10^{-20} \ kg.m/s)(2.998 \times 10^8 \ m/s) = 3.298 \times 10^{-12}$ J, 

so, K = $\sqrt{(pc)^2+(mc^2)^2}-mc^2$

K = $\sqrt{(3.30 \times 10^{-12})^2+(1.50 \times 10^{-10})^2}-1.501 \times 10^{-10}$

K = $3.622 \times 10^{-14} \ J$ = 0.23 MeV

(c) The result of part (b), about $2 \times 10^5$ eV, is many orders of magnitude larger than the potential energy of an electron in a hydrogen atom.

#39.81

Electron Energy in a Nucleus. The radii of atomic nuclei are of the order of 5,0 $\times$ $10^{-15}$ m

(a) Estimate the minimum uncertainty in the momentum of an electron if it is confined within a nucleus. 

(b) Take this uncertainty in momentum to be an estimate of the magnitude of the momentum. Use the relativistic relationship between energy and momentum, Eq. (37.39), to obtain an estimate of the kinetic energy of an electron confined within a nucleus. 

(c) Compare the energy calculated in part (b) to the magnitude of the Coulomb potential energy of a proton and an electron separated by 5,0 $\times$ $10^{-15}$ m. On the basis of your result, could there be electrons within the nucleus? (Note: It is interesting to compare this result to that of Problem 39.80.)

Answer:

(a) Apply the Heisenberg Uncertainty Principle. Consider only one component of position and momentum.

$\Delta x \Delta p_x$ = $\frac{\hbar}{2}$

Estimate $\Delta x$ as $\Delta x \approx 5.0 \times 10^{-15}$ m. 

Then the minimum allowed $\Delta p_x$ is 

$\Delta p_x \approx \frac{\hbar}{2\Delta x}=\frac{1.055 \times 10^{-34} \ J.s}{2(5.0 \times 10^{-15} / m)}$

$\Delta p_x$ = 1.1 $\times 10^{-20}$ kg.m/s

(b) Assume $p \approx 1.1 \times 10^{-20}$ kg.m/s. Use Eq. (37.39) to calculate E, and then

K = E - $mc^2$.

E = $\sqrt{(pc)^2+(mc^2)^2}$

 $mc^2 = (9.109 \times 10^{-31} \ kg)(2.998 \times 10^8 \ m/s)^2 = 8.187 \times 10^{-14}$ J  

and pc = $(1.1 \times 10^{-20} \ kg.m/s)(2.998 \times 10^8 \ m/s) = 3.298 \times 10^{-12}$ J, 

E = $\sqrt{(3.298 \times 10^{-12} \ J)^2+(8.187 \times 10^{-14})^2}$

E = $3.299 \times 10^{-12}$ J

So, K = $3.299 \times 10^{-12}$ J - 8.187 \times 10^{-14}$ J  = 3.216 $\times 10^{-12}$ J

K = 20.1 MeV

(c) The Coulomb potential energy for a pair of point charges is given by $U = -\frac{ke^2}{r}$. The proton has charge +e and the electron has charge –e.

So, $U = -\frac{(8.988 \times 10^{-9} N.m^2/C^2)(1.602 \times 10^{-19} \ C)^2}{5.0 \times 10^{-15} \ m}$

U = $-4.6 \times 10^{-14}$ J = $-0.29$ MeV

The kinetic energy of the electron required by the uncertainty principle would be much larger than the magnitude of the negative Coulomb potential energy. The total energy of the electron would be large and positive and the electron could not be bound within the nucleus. 

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