The radius of a hypothetical nucleus -atomic number-79- is about 7times10-15- m- Assuming that charge distribution is uniform- the electric field the surface of the nucleus is -2-9times10-21-2-32times10-21-4-64times10-21-0

Question

Toppr Admin · Accepted Answer

The expression of electric field at surface of sphere of radius a is given by, $E = \frac{k q}{a^{2}}$
Here, charge $q = 79 e = 79 \times (1.6 \times 10^{- 19})$ and $a = 7 \times 10^{- 15}$

So, $E = \frac{(9 \times 10^{9}) \times (79 \times 1.6 \times 10^{- 19})}{(7 \times 10^{- 15})^{2}} = 2.32 \times 10^{21} N / C$

The radius of a hypothetical nucleus (atomic number=79) is about 7×10−15 m. Assuming that charge distribution is uniform, the electric field at the surface of the nucleus is :2.9×10−212.32×10214.64×10210

The expression of electric field at surface of sphere of radius a is given by, E=kqa2Here, charge q=79e=79×(1.6×10−19) and a=7×10−15So, E=(9×109)×(79×1.6×10−19)(7×10−15)2=2.32×1021N/C

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 :
$2.9 \times 10^{- 21}$
$2.32 \times 10^{21}$
$4.64 \times 10^{21}$
$0$