SATHEE: Electrostatics Question 490

Electrostatics Question 490

Question: The radius of nucleus of silver (atomic number = 47) is $3.4 \times 10^{- 14} m$ . The electric potential on the surface of nucleus is $(e = 1.6 \times 10^{- 19} C)$ [Pb. PET 2003]

Options:

A) $1.99 \times 10^{6} v o l t$

B) $2.9 \times 10^{6} v o l t$

C) $4.99 \times 10^{6} v o l t$

D) $0.99 \times 10^{6} v o l t$

Show Answer

Answer:

Correct Answer: A

Solution:

$V = \frac{1}{4 π ε_{0}} . \frac{(Z e)}{r} = 9 \times 10^{9} \times \frac{47 \times 1.6 \times 10^{- 19}}{3.4 \times 10^{- 14}} = 1.99 \times 10^{6} V$

Electrostatics Question 490

Question: The radius of nucleus of silver (atomic number = 47) is $3.4 \times 10^{- 14} m$ . The electric potential on the surface of nucleus is $(e = 1.6 \times 10^{- 19} C)$ [Pb. PET 2003]

Options:

Answer:

Solution:

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Electrostatics Question 490

Question: The radius of nucleus of silver (atomic number = 47) is 3.4×10−14m . The electric potential on the surface of nucleus is (e=1.6×10−19C) [Pb. PET 2003]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: The radius of nucleus of silver (atomic number = 47) is $3.4 \times 10^{- 14} m$ . The electric potential on the surface of nucleus is $(e = 1.6 \times 10^{- 19} C)$ [Pb. PET 2003]