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

Asked in JEE Main 27th Jan 2nd Shift 2024 · Distance of closest approach

Answer: 8

Step-by-step solution

r=9×10⁻¹⁵ m, q=50×1.6×10⁻¹⁹=8×10⁻¹⁸ C

V=(kq)/r=(9×10⁹×8×10⁻¹⁸)/(9×10⁻¹⁵)=8×10⁶ V

→ 8

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