The radius of nucleus of silver (atomic number =47 ) is 3.4 × 10-14 m . The electric potential on the surface of nucleus is (e=1.6 × 10-19 C )

Question

The radius of nucleus of silver (atomic number =47 ) is 3.4 × 10-14 m . The electric potential on the surface of nucleus is (e=1.6 × 10-19 C ) (A) 1.99 × 106 volt (B) 2.9 × 106 volt (C) 4.99 × 106 volt (D) 0.99 × 106 volt

Tardigrade Admin · Accepted Answer

V=(1/4 π ε0) ⋅ ((Z e)/r) =9 × 109 × (47 × 1.6 × 10-19/3.4 × 10-14) =1.99 × 106 V

Q. The radius of nucleus of silver (atomic number $= 47$ ) is $3.4 \times 1 0^{- 14} m .$ The electric potential on the surface of nucleus is $(e = 1.6 \times 1 0^{- 19} C)$

$1.99 \times 1 0^{6}$ volt

$2.9 \times 1 0^{6}$ volt

$4.99 \times 1 0^{6}$ volt

$0.99 \times 1 0^{6}$ volt

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

Q. 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)

1.99×106 volt

2.9×106 volt

4.99×106 volt

0.99×106 volt