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The electric potential at the surface of an atomic nucleus (Z = 50) of radius 9\times 10^5 V is 

  • Option 1)

    80 V

  • Option 2)

    8\times 10^6 V

  • Option 3)

    9 V

  • Option 4)

    9\times 10^5 V

 

Answers (1)

As we have learnt,

 

Potential energy Per unit charge -

V =\frac{W}{Q}=\frac{U}{Q}

- wherein

S.I unit is \frac{J}{C}.

 

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

 


Option 1)

80 V

Option 2)

8\times 10^6 V

Option 3)

9 V

Option 4)

9\times 10^5 V

Posted by

Vakul

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