Question

Bodies P, Q, R, S are labelled as having charges $Q_P=5\times10^{-19}\text{ C}$, $Q_Q=7\times10^{-19}\text{ C}$, $Q_R=1\times10^{-19}\text{ C}$, $Q_S=8\times10^{-19}\text{ C}$ respectively. Select the body having the correct charge. [Given electronic charge $e = 6\times10^{-19}\text{ C}$]}

• A. $Q_P=0.5\times10^{-19}\text{ C}$
• B. $Q_S=4.8\times10^{-19}\text{ C}$
• C. $Q_R=2.1\times10^{-19}\text{ C}$
• D. $Q_Q=0.7\times10^{-19}\text{ C}$

Hint:

To solve quantization problems quickly, ignore the $10^{-19}$ exponent factor completely and look for a value that is perfectly divisible by $1.6$. Since $1.6 \times 3 = 4.8$, Body S stands out immediately as the only valid option!

Answer 1

The Correct Option is B

Understanding the Concept: According to the principle of quantization of electric charge, the total charge ($Q$) possessed by any physical body must always be an integral multiple of the basic fundamental electronic charge unit ($e$): \[ Q = n \cdot e \] where $n$ must be a whole integer ($n = \pm 1, \pm 2, \pm 3, \dots$). A net stable charge value cannot exist as a fractional component of an electron.
Step 1: Test each body by calculating its equivalent electron count ($n$).
Given baseline fundamental unit: $e = 1.6\times10^{-19}\text{ C}$. Let's divide each charge option by $e$:
Body P: $n = \frac{0.5\times10^{-19}}{1.6\times10^{-19}} = \frac{0.5}{1.6} = 0.3125$ (Not an integer)
Body Q: $n = \frac{0.7\times10^{-19}}{1.6\times10^{-19}} = \frac{0.7}{1.6} = 0.4375$ (Not an integer)
Body R: $n = \frac{2.1\times10^{-19}}{1.6\times10^{-19}} = \frac{2.1}{1.6} = 1.3125$ (Not an integer)
Body S: $n = \frac{4.8\times10^{-19}}{1.6\times10^{-19}} = \frac{4.8}{1.6} = 3$ (Perfect Integer!)

Step 2: Identify the physically possible state.
Since body S corresponds to an exact integral value ($n = 3$ excess or deficit electrons), it is the only physically possible charge distribution among the choices given.