The electric potential difference ($\Delta V_e$) required to stop an electron moving with an initial speed $v_0 = 2.10 \times 10^7 \, \text{m/s}$ can be calculated using the principle of conservation of energy. Specifically, the kinetic energy of the electron is converted into electrical potential energy.

The kinetic energy ($K.E.$) of the electron is given by:

$K.E. = \frac{1}{2} m_e v_0^2$

where:

• $m_e = 9.109 \times 10^{-31} \, \text{kg}$ is the mass of the electron,
• $v_0 = 2.10 \times 10^7 \, \text{m/s}$ is the initial speed of the electron.

The electrical potential energy change is:

$\Delta U = e \Delta V_e$

where:

• $e = 1.602 \times 10^{-19} \, \text{C}$ is the elementary charge (the charge of the electron),
• $\Delta V_e$ is the potential difference.

Setting the kinetic energy equal to the change in electrical potential energy:

$\frac{1}{2} m_e v_0^2 = e \Delta V_e$

Solving for $\Delta V_e$:

$\Delta V_e = \frac{\frac{1}{2} m_e v_0^2}{e}$

Now let's calculate $\Delta V_e$.The electric potential difference ($\Delta V_e$) required to stop the electron is approximately $1.25 \, \text{kV}$.

Would you like more details or have any questions?

Here are 5 related questions you might explore:

1. How does the stopping potential vary with the speed of the electron?
2. What is the relationship between electric potential and kinetic energy?
3. How does the mass of a particle affect the stopping potential?
4. What happens if a proton is used instead of an electron in this scenario?
5. How do we measure stopping potential experimentally?

Tip: The stopping potential depends only on the speed of the particle, not its charge, as long as it's uniformly charged.