Question

Under the action of a given coulombic force the acceleration of an electron is $2.5 \times 10^{22}\text{ m s}^{-2}$. Then the magnitude of the acceleration of a proton under the action of same force is nearly

• A. $1.6 \times 10^{-19}\text{ m s}^{-2}$
• B. $9.1 \times 10^{31}\text{ m s}^{-2}$
• C. $1.5 \times 10^{19}\text{ m s}^{-2}$
• D. $1.6 \times 10^{27}\text{ m s}^{-2}$

Hint:

Since both particles feel a force of the same size, use the second law of motion, F = m times a, to see that acceleration is inversely proportional to mass. Recall the standard mass values of an electron and a proton, and keep careful track of the powers of ten when you divide.

Answer 1

The Correct Option is C

Concept: According to Newton's Second Law of Motion, the acceleration ($a$) produced by a net force ($F$) acting on an object is inversely proportional to its mass ($m$): \[ F = m \cdot a \quad \implies \quad a = \frac{F}{m} \] When two different particles experience the exact same magnitude of force ($F_e = F_p = F$), their accelerations are inversely proportional to their respective rest masses: \[ m_e \cdot a_e = m_p \cdot a_p \] Rearranging this relationship allows us to find the unknown acceleration of the proton ($a_p$): \[ a_p = a_e \cdot \left( \frac{m_e}{m_p} \right) \]

Step 1: Identify the fundamental constant values and given metrics.

• Acceleration of the electron ($a_e$) = $2.5 \times 10^{22}\text{ m s}^{-2}$
• Rest mass of an electron ($m_e$) $\approx 9.1 \times 10^{-31}\text{ kg}$
• Rest mass of a proton ($m_p$) $\approx 1.67 \times 10^{-27}\text{ kg}$

Step 2: Set up the ratio equation for the proton's acceleration.
Substituting the metrics into our ratio formula: \[ a_p = (2.5 \times 10^{22}) \times \frac{9.1 \times 10^{-31}}{1.67 \times 10^{-27}} \] Let us group the numerical coefficients and the power terms together to make simplification clean: \[ a_p = \left( \frac{2.5 \times 9.1}{1.67} \right) \times \frac{10^{22} \times 10^{-31}}{10^{-27}} \]

Step 3: Simplify the exponents and perform the final arithmetic calculation.
First, simplify the powers of 10 in the numerator: \[ 10^{22} \times 10^{-31} = 10^{-9} \] Now divide by the denominator's power base using exponent subtraction rules ($10^{-9} / 10^{-27} = 10^{-9 - (-27)} = 10^{18}$): \[ \text{Total Exponent} = 10^{18} \] Next, compute the numerical decimal fraction: \[ 2.5 \times 9.1 = 22.75 \] \[ \frac{22.75}{1.67} \approx 13.62 \] Assemble the parts back together: \[ a_p \approx 13.62 \times 10^{18}\text{ m s}^{-2} \] Converting the final value into standard scientific notation by moving the decimal place one step to the left: \[ a_p \approx 1.362 \times 10^{19}\text{ m s}^{-2} \approx 1.5 \times 10^{19}\text{ m s}^{-2} \] Therefore, the magnitude of the acceleration of the proton under the action of the same force is nearly $1.5 \times 10^{19}\text{ m s}^{-2}$.

Answer 2

Concept:

• When two different particles feel forces of the same size, the second law of motion, $F = m \times a$, shows their accelerations must be inversely proportional to their masses, so only the ratio of the two masses is needed, not the actual force value.
• Knowing the standard rest mass ratio between a proton and an electron by heart turns a messy scientific notation calculation into a single division.

Step 1: Set up the inverse relation between the two accelerations.
Since $F_{electron} = F_{proton} = F$, we get $F = m_e \times a_e = m_p \times a_p$, so $a_p = a_e \times \dfrac{m_e}{m_p}$.

Step 2: Recall the standard mass ratio.
The rest mass of a proton is close to 1836 times the rest mass of an electron, so $\dfrac{m_p}{m_e} \approx 1836$.

Step 3: Divide to get the proton acceleration.
$a_p = \dfrac{2.5 \times 10^{22}}{1836}$. Breaking the division into an easy form, $\dfrac{2.5}{1836} \approx 0.001362$, so $a_p \approx 0.001362 \times 10^{22} = 1.362 \times 10^{19}\text{ m s}^{-2}$.

Step 4: Match with the closest option.
$1.362 \times 10^{19}$ lands in the same order of magnitude bracket as the option $1.5 \times 10^{19}\text{ m s}^{-2}$, while every other listed option differs by ten or more orders of magnitude, so it cannot be a rounding candidate.

Final Answer: $1.5 \times 10^{19}\text{ m s}^{-2}$ nearly.