Compute the electric potential the surface of the nucleus of a silver atom- The radius of the nucleus is 3-4 times -10- - 14-m and the atomic number of silver is 47-

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PE-x2212-kze2rn-P-x2212-kzern-x2212-1-xD7-47-xD7-1-6-xD7-10-x2212-194-x3C0-x3B5-xD7-3-4-xD7-10-x2212-14-9-xD7-109-xD7-47-xD7-1-6-xD7-10-x2212-193-4-xD7-10-x2212-14-199-05-xD7-10-x2212-10-4-199-05-xD7-104

Compute the electric potential at the surface of the nucleus of a silver atom. The radius of the nucleus is $3.4 \times 10^{- 14} m$ and the atomic number of silver is $47.$

$P E = (\frac{- k z e^{2}}{r_{n}})$
$P = (\frac{- k z e}{r_{n}})$
$= \frac{- 1 \times 47 \times 1.6 \times 10^{- 19}}{4 π ε \times 3.4 \times 10^{- 14}}$
$= \frac{9 \times 10^{9} \times 47 \times 1.6 \times 10^{- 19}}{3.4 \times 10^{- 14}}$
$= 199.05 \times 10^{- 10 + 4}$
$= 199.05 \times 10^{4}$

Compute the electric potential at the surface of the nucleus of a silver atom. The radius of the nucleus is 3.4×10−14m and the atomic number of silver is 47.

PE=(−kze2rn)P=(−kzern)=−1×47×1.6×10−194πε×3.4×10−14=9×109×47×1.6×10−193.4×10−14=199.05×10−10+4=199.05×104

Compute the electric potential at the surface of the nucleus of a silver atom. The radius of the nucleus is $3.4 \times 10^{- 14} m$ and the atomic number of silver is $47.$

$P E = (\frac{- k z e^{2}}{r_{n}})$
$P = (\frac{- k z e}{r_{n}})$
$= \frac{- 1 \times 47 \times 1.6 \times 10^{- 19}}{4 π ε \times 3.4 \times 10^{- 14}}$
$= \frac{9 \times 10^{9} \times 47 \times 1.6 \times 10^{- 19}}{3.4 \times 10^{- 14}}$
$= 199.05 \times 10^{- 10 + 4}$
$= 199.05 \times 10^{4}$