The electric potential at the surface of an atomic nucleus (z=50) of radius 9 × 10⁻¹³ cm is __________ × 10⁶ V.

Solution

Physics sir · 3 steps

Outside or on its surface, a charged sphere acts like a point charge at its centre, so use V = kQ/R.

1. Charge on nucleus. The nucleus holds Z = 50 protons, so Q = 50 × 1.6 × 10⁻¹⁹ = 8 × 10⁻¹⁸ C.
2. Radius in metres. R = 9 × 10⁻¹³ cm = 9 × 10⁻¹⁵ m. The constant k = 9 × 10⁹ N m² C⁻² is in SI units, so R must be in metres too.

   V = kQ/R = (9 × 10⁹ × 8 × 10⁻¹⁸)/(9 × 10⁻¹⁵) = 8 × 10⁶ V

3. So. The potential at the surface is 8 × 10⁶ V, so the answer is 8.

The common slip: 0.08. Using the radius as 9 × 10⁻¹³ m without converting from cm gives 8 × 10⁴ V, which is 0.08 in units of 10⁶ V, not 8.

The idea to keep. For a spherical charge, the potential at the surface or outside is kQ/R, the same as for a point charge. Convert every length to metres before using k = 9 × 10⁹.

Answer from the NTA final answer key. Solution by PrepWiser.