Question

Under the action of a given coulombic force the acceleration of an electron is 2.5 $\times$ 10²² m s⁻². Then the magnitude of the acceleration of a proton under the action of same force is nearly

• A. 1.6 $\times$ 10⁻¹⁹ m s⁻²
• B. 9.1 $\times$ 10³¹ m s⁻²
• C. 1.5 $\times$ 10¹⁹ m s⁻²
• D. 1.6 $\times$ 10²⁷ m s⁻²

Hint:

A proton is roughly 1836 times heavier than an electron ($m_p \approx 1836 \cdot m_e$). You can get the answer quickly by dividing the given electron acceleration directly by 1800: $$\frac{2.5 \times 10^{22}}{1836} \approx 1.36 \times 10^{19}\text{ m s}^{-2}$$

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
According to Newton's Second Law of Motion, acceleration is inversely proportional to the mass of an object when a constant force is applied. Because a proton is significantly heavier than an electron, it will experience a much smaller acceleration when subjected to the exact same coulombic force.

Step 2: Key Formula or Approach:
$$F = m \cdot a \implies a = \frac{F}{m}$$ Since the force ($F$) acting on both particles is identical: $$m_e \cdot a_e = m_p \cdot a_p \implies a_p = a_e \times \frac{m_e}{m_p}$$ Where: $a_e = 2.5 \times 10^{22}\text{ m s}^{-2}$ Mass of an electron ($m_e$) $\approx 9.1 \times 10^{-31}\text{ kg}$ Mass of a proton ($m_p$) $\approx 1.67 \times 10^{-27}\text{ kg}$

Step 3: Detailed Explanation:
Substitute the values of the masses and the given electron acceleration into our ratio equation: \[ a_p = (2.5 \times 10^{22}) \times \frac{9.1 \times 10^{-31}}{1.67 \times 10^{-27}} \] Combine the base decimal terms and the exponential powers of 10 separately: \[ a_p = \left( \frac{2.5 \times 9.1}{1.67} \right) \times 10^{22 - 31 - (-27)} \] \[ a_p = \left( \frac{22.75}{1.67} \right) \times 10^{18} \] \[ a_p \approx 13.62 \times 10^{18}\text{ m s}^{-2} = 1.362 \times 10^{19}\text{ m s}^{-2} \] Rounding to the closest value listed among the multiple-choice options, it is approximately 1.5 $\times$ 10¹⁹ m s⁻².

Step 4: Final Answer:
The magnitude of the acceleration of the proton is nearly 1.5 $\times$ 10¹⁹ m s⁻².