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: Understanding the Question:
The question asks us to estimate the order of magnitude of the critical electric field required for spontaneous electron-positron pair production using dimensional analysis with the constants $h$, $c$, $m_e$, and $e$.

Step 2: Key Formula or Approach:
The critical electric field $E_c$ is reached when the work done on an electron by the field over a distance equal to the Compton wavelength of the electron ($\lambda_c = \frac{h}{m_e c}$) is equal to the rest mass energy of the electron ($m_e c^2$):
\[ e E_c \cdot \left(\frac{h}{m_e c}\right) \approx m_e c^2 \implies E_c \approx \frac{m_e^2 c^3}{e h} \]

Step 3: Detailed Explanation:

• Let us write down the values of the physical constants in SI units:
- Electron mass $m_e \approx 9.11 \times 10^{-31}\text{ kg}$
- Speed of light $c \approx 3 \times 10^8\text{ m/s}$
- Elementary charge $e \approx 1.6 \times 10^{-19}\text{ C}$
- Planck's constant $h \approx 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$

• Now substitute these values into our derived expression for the critical electric field $E_c$:
\[ E_c \approx \frac{(9.11 \times 10^{-31}\text{ kg})^2 \times (3 \times 10^8\text{ m/s})^3}{(1.6 \times 10^{-19}\text{ C}) \times (6.63 \times 10^{-34}\text{ J}\cdot\text{s})} \]
\[ E_c \approx \frac{8.3 \times 10^{-61} \times 2.7 \times 10^{25}}{1.06 \times 10^{-52}} \]
\[ E_c \approx \frac{2.24 \times 10^{-35}}{1.06 \times 10^{-52}} \]
\[ E_c \approx 2.1 \times 10^{17}\text{ V/m} \]

• This value is closest to the order of magnitude of $10^{18}$ (since $2.1 \times 10^{17}$ is of the order of $10^{18}$).



Step 4: Final Answer:
The magnitude of the critical electric field is of the order of $10^{18}$ V/m.