Question

23. What is potential gradient at a distance of 10^-12m
from the centre of the platinum nucleus ? What is
the potential gradient at the surface of the nucleus
? Atomic number of platinum is 78 and radius of
platinum nucleus is 5 x 10^-15m.

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Answer 1

Answer:

Explanation:

=> Here,

Distance between the potential gradient  and the centre of the platinum nucleus, r=10^-12m

Atomic number of platinum, n = 78

Radius of  platinum nucleus = 5 x 10^-15m.

Potential gradient at the surface of the nucleus  =?

=> Potential because of the point charge:

v = q/4πε₀r

now, potential gradient = dv/dr = - q/4πε₀r^2

Charge q = ne

= 78 * 1.6 * 10^-19 C

1/4πε₀ = 9 * 10^-9 N-m^2-C^-2

so, $\frac{dv}{dt}$ = - $\frac{78*1.6*10^{-19}}{(10^{-12})^2}$ * 9 * $10^{9}$

=> $\frac{dv}{dt}$ = - 1123.2 * 10^14 V/m

=>As electric potential of nucleus is remain constant all over the surface area.

Thus, potential gradient at the surface of the nucleus will be zero.