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 $47.$

Open in App

Solution

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

Suggest Corrections

Similar questions

Q. The electric potential at the surface of an atomic nucleus

(Z = 50)

of radius

9 \times 10^{- 15} m

Q. Determine the electric potential at the surface of the gold nucleus. The radius is

6.6 \times 10^{- 5} m

and the atomic number

Z = 79

. Given charge on a proton

= 1.6 \times 10^{- 19} C .

Q. An alpha particle after passing through a potential difference of

2 \times 10^{6} v o l t

falls on a silver foil. The atomic number of silver is

47

. Calculate

(I)

the

K . E

of the alpha-particle at the time of falling on the foil.

(I I) K . E

of the alpha-particle at a distance of

5 \times 10^{- 14} m

from the nucleus,

(I I I)

the shortest distance from the nucleus of silver to which the alpha-particle reaches.

Q. The radius of a hypothetical nucleus (atomic number

= 79

) is about

7 \times 10^{- 15}

m. Assuming that charge distribution is uniform, the electric field at the surface of the nucleus is :

An electron is circulating around the nucleus of the hydrogen atom in a circular orbit of radius 5.3×10^-11 m calculate the electric potential at this radius and electric potential energy of the atom in EV.

Join BYJU'S Learning Program

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 $47.$

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