The electric potential at the surface of an atomic nucleus (Z = 5… — Solution | JEE Physics

NVAEasyJEE 2024Electric Potential & Potential Energy

The electric potential at the surface of an atomic nucleus ( $Z = 50$ ) of radius $9\times 10^{-13} \, \text{cm}$ is $\times 10^6 \, \text{V}$ :

Answer

Correct answer:8

Standard Method

Given: $Z = 50$ , radius $R = 9 \times 10^{-13} \, \text{cm} = 9 \times 10^{-15} \, \text{m}$ .

Find: The electric potential at the surface of the nucleus in the form $\_\_\_ \times 10^6 \, \text{V}$ .

For a nucleus, the surface potential is

V = \frac{kQ}{R} = \frac{kZe}{R}

Using $k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2$ and $e = 1.6 \times 10^{-19} \, \text{C}$ ,

V = \frac{9 \times 10^9 \times 50 \times 1.6 \times 10^{-19}}{9 \times 10^{-15}}

Now simplify:

V = 8 \times 10^6 \, \text{V}

Therefore, the required numerical value is $8$ .

Direct Substitution

Given: $Z = 50$ and $R = 9 \times 10^{-15} \, \text{m}$ .

Find: The numerical coefficient in the potential.

Use the direct relation

V = \frac{kZe}{R}

Substitute the values immediately:

V = \frac{9 \times 10^9 \times 50 \times 1.6 \times 10^{-19}}{9 \times 10^{-15}} = 8 \times 10^6 \, \text{V}

So, the correct answer is $8$ .

Using the radius in cm instead of converting it to m is incorrect because Coulomb's constant is in SI units. Always convert $9 \times 10^{-13} \, \text{cm}$ to $9 \times 10^{-15} \, \text{m}$ first.
Taking the nuclear charge as $e$ instead of $Ze$ is wrong because the nucleus has total charge $Q = Ze$ . Here, use $Q = 50e$ , not just $e$ .
Dropping the power of ten during simplification leads to a wrong answer. Track the exponents carefully while evaluating $\frac{10^9 \times 10^{-19}}{10^{-15}} = 10^5$ and then include the numerical factors correctly to get $8 \times 10^6$ .

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