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Physics Question on Nuclei

The electric potential at the surface of an atomic nucleus (z=50)(z = 50) of radius 9×10−139 \times 10^{-13} cm is \\_\\_\\_\\_\\_\\_\\_ $$\times 10^{6} V.

Answer

The electric potential at the surface of a nucleus is given by:

V=kQR=kZeR,V = \frac{kQ}{R} = \frac{kZe}{R},

where:
- k=9×109 N\cdotpm2/C2k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2,
- Z=50Z = 50,
- e=1.6×10−19 Ce = 1.6 \times 10^{-19} \, \text{C},
- R=9×10−13 cm=9×10−15 mR = 9 \times 10^{-13} \, \text{cm} = 9 \times 10^{-15} \, \text{m}.

Substituting the values:

V=9×109×50×1.6×10−199×10−15V = \frac{9 \times 10^9 \times 50 \times 1.6 \times 10^{-19}}{9 \times 10^{-15}}

Simplify:

V=8×106 V.V = 8 \times 10^6 \, \text{V}.
The Correct answer is: 8