Question

The critical electric field required to produce electron-positron pairs depends on the physical constants $h$, $c$, $m_e$ and $e$. Use dimensional analysis and assume that the dimensionless coefficient is of order one. The magnitude of the critical electric field, in SI units, is of the order

• A. $10^{18}$
• B. $10^{21}$
• C. $10^{24}$
• D. $10^{15}$

Hint:

Using the reduced Planck's constant $\hbar = \frac{h}{2\pi} \approx 1.05 \times 10^{-34}\text{ J}\cdot\text{s}$ gives:
\[ E_c = \frac{m_e^2 c^3}{e \hbar} \approx 1.3 \times 10^{18}\text{ V/m} \]
which directly confirms the order of magnitude of $10^{18}$.

Answer 1

The Correct Option is A

Step 1: Understand the physics behind the field.
A field can rip an electron-positron pair from the vacuum when it does enough work over a quantum length scale to supply the energy $m_e c^2$.

Step 2: Identify the length scale.
The natural quantum length here is the reduced Compton wavelength, of order $\lambda \sim h / (m_e c)$.

Step 3: Set up the work-energy balance.
The work done by the field over that length should match the rest energy needed, \[ e E_c \, \lambda \sim m_e c^2. \]

Step 4: Solve for the critical field.
\[ E_c \sim \frac{m_e c^2}{e \, \lambda} = \frac{m_e c^2 \cdot m_e c}{e \, h} = \frac{m_e^2 c^3}{e \, h}. \]

Step 5: Plug in the constants.
Using $m_e \approx 9.1 \times 10^{-31}$, $c \approx 3 \times 10^{8}$, $e \approx 1.6 \times 10^{-19}$, $h \approx 6.6 \times 10^{-34}$, the numerator $m_e^2 c^3 \approx (8.3 \times 10^{-61})(2.7 \times 10^{25}) \approx 2.2 \times 10^{-35}$, and the denominator $e h \approx 1.06 \times 10^{-52}$.

Step 6: Take the ratio.
\[ E_c \approx \frac{2.2 \times 10^{-35}}{1.06 \times 10^{-52}} \approx 2 \times 10^{17} \sim 10^{18}\ \text{V/m}. \] \[ \boxed{E_c \sim 10^{18}\ \text{V/m}} \]