A tin nucleus has a charge of +50c. If the proton is at a distance 10-22 m from the nucleus, then the potential V at this position is (charge on the proton=1.6 x 10^-19C):
Correct answer: C. 7.2 x 10^14 volt
- A. 14.4 x 10^4 volt
- B. 7.2 x 10^8 volt
- C. 7.2 x 10^14 volt
- D. 14.4 x 10^8 volt
Explanation
Option C is correct Number of protons in tin nucleus = 50 Total charge = 50 X 1.66X10-19 Potential at that point = k x (net charge ) / distance V = 9 x 109 X 50 X 1.66X10-19 / 1X10-22 V = 7.4 X 1014
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