Chapter 6, Problem 52QP

In the beginning of the twentieth century, some scientists thought that a nucleus may contain both electrons and protons. Use the Heisenberg uncertainty principle to show that an electron cannot ha confined within a nucleus. Repeat the calculation for a proton Comment on your results Assume the radius of a nucleus. to be $\text{1}\text{.0 }\times {\text{ 10}}^{\text{–13}}\text{m}\text{.}$ The masses of an electron and a proton are $\text{ 9}\text{.109 }\times {\text{ 10}}^{\text{–31}}\text{ kg and 1}\text{.673 }\times {\text{ 10}}^{\text{–27}}\text{ kg}$ respectively (Hints: Treat the radius of the nucleus as the uncertainty in position)

Interpretation Introduction

Interpretation:

The statement, “An electron cannot be confined within the nucleus”, and the statement, “A proton cannot be confined within the nucleus”, are to be proved by using the Heisenberguncertaintyprinciple.

Concept Introduction:

According to the Heisenberg uncertainty principle, the product of the uncertainty in position and momentum of a particle cannot be less than $\frac{h}{4\pi }$. It is expressed as follows:

$\Delta x\cdot \Delta p\ge \frac{h}{4\pi }$.

Here, $\Delta x$ is uncertainty in position, $\Delta p$ is uncertainty in momentum, and $h$ is Planck’s constant $\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)$.

As momentum is the product of mass and velocity, the equation of Heisenberg uncertainty principle can also be written as follows:

$\Delta x\cdot m\Delta u\ge \frac{h}{4\pi }$

Here, $\Delta x$ is uncertainty in the position of theparticle, $m$ is the mass of the particle, $\Delta u$ is uncertainty in velocity, and $h$ is Planck’s constant $\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)$.

Answer to Problem 52QP

Solution: The calculated speed of anelectron is higher than the speed of light, so it cannot be confined within the nucleus. The uncertainty in velocity of the electron is $5.8{\text{×10}}^{\text{10}}\text{ m/s}$.

The value of uncertainty of a proton is less than the speed of light, so a proton cannot be confined within the nucleus. The uncertainty in velocity of the proton is $3.2{\text{×10}}^7\text{ m/s}$.

Explanation of Solution

Given information:

The mass of an electron $m_e$ is $9.109\times {10}^{-31}\text{ kg}$.

The mass of a proton $m_p$ is $1.675\times {10}^{-27}\text{ kg}$.

The radius of the nucleus is the uncertainty in position that is $1.0\times {10}^{-15}\text{ m}$.

The uncertainty in velocity of an electron can be evaluated as follows:

$\Delta x\cdot m_e\Delta u=\frac{\text{h}}{4\pi }$

Substitute $6.63\times {10}^{-34}\text{ J}\cdot \text{s}$ for $\text{h}$, $1.0\times {10}^{-15}\text{ m}$ for $\Delta x$, and $9.109\times {10}^{-31}\text{ kg}$ for $m_e$,

$\begin{array}{c}\left(1.0\times {10}^{-15}\text{ m}\right)\left(9.109\times {10}^{-31}\text{ kg}\right)\Delta u=\frac{6.63\times {10}^{-34}\text{ J}\cdot \text{s}}{4\pi }\\ \Delta u=\frac{6.63\times {10}^{-34}\cancel{\text{J}}\cdot \cancel{\text{s}}}{4\pi \left(1.0\times {10}^{-15}\cancel{\text{m}}\right)\left(9.109\times {10}^{-31}\cancel{\text{kg}}\right)}\left(\frac{1.0\cancel{\text{kg}}\cancel{\text{m}}\text{.m }\cancel{{\text{s}}^{-1}}{\text{.s}}^{-1}}{1.0\cancel{\text{J}}}\right)\\ =5.8{\text{×10}}^{\text{10}}\text{ m/s}\end{array}$

Therefore, the uncertainty invelocity of the electron is $5.8{\text{×10}}^{\text{10}}\text{ m/s}$.

This value for the uncertainty is impossible as it exceeds the speed of light. If the electron exists in the nucleus, then it must travel with a speed of $5.8{\text{×10}}^{\text{10}}\text{ m/s}$. It contradicts the theory of relativity. If an object does not have any mass, then it can travel faster than the speed of light $\text{3}{\text{.0×10}}^{\text{8}}\text{ m/s}$, but electrons have mass. Therefore, it cannot travel faster than the speed of light. Hence, it cannot be confined within the nucleus.

The uncertainty in velocity of the proton can be evaluated as follows:

$\Delta x\cdot m_p\Delta u=\frac{\text{h}}{4\pi }$

Substitute $6.63\times {10}^{-34}\text{ J}\cdot \text{s}$ for $\text{h}$, $1.0\times {10}^{-15}\text{ m}$ for $\Delta x$, and $1.675\times {10}^{-27}\text{ kg}$ for $m_p$,

$\begin{array}{c}\left(1.0\times {10}^{-15}\text{ m}\right)\left(1.675\times {10}^{-27}\text{ kg}\right)\Delta u=\frac{6.63\times {10}^{-34}\text{ J}\cdot \text{s}}{4\pi }\\ \Delta u=\frac{6.63\times {10}^{-34}\cancel{\text{J}}\cdot \cancel{\text{s}}}{4\pi \left(1.0\times {10}^{-15}\cancel{\text{m}}\right)\left(1.673\times {10}^{-27}\text{ kg}\right)}\left(\frac{1.0\text{ kg }\cancel{\text{m}}{\text{.m s}}^{-1}\text{.}\cancel{{\text{s}}^{-1}}}{1.0\cancel{\text{J}}}\right)\\ =3.2{\text{×10}}^7\text{ m/s}\end{array}$

Therefore, the uncertainty invelocity of the proton is $3.2{\text{×10}}^7\text{ m/s}$.

The value of uncertainty of proton is less than the speed of light, so a proton cannot be confined within the nucleus.

Conclusion

When the electron exists in the nucleus, it must travel with a speed of $5.8{\text{×10}}^{\text{10}}\text{ m/s}$. It contradicts the theory of relativity. The uncertainty in velocity of the proton is $3.2{\text{×10}}^7\text{ m/s}$. Hence, electrons and protons cannot be confined within the nucleus.