What is potential gradient at a distance of `10^(-12) m` from the centre of the platinum nucleas ? What is the potential gradient at the surface of the nucleas ? Atomic number fo platinum is 78 and radius of platinum nucleas is `5xx10^(-15) m`.

Answer

`1.23xx10^(17) Vm^(-1) ; 4.5xx10^(21) Vm^(-1)`

Explanation

Here, `q = Ze = 78xx1.6xx10^(-19)C`
Potential gradient at a point is numberically equal to electric field at that point, i.e.,
`(dV)/(dr) = E = (q)/(4pi in_(0) r^(2))`
At `r = 10^(-12)m`,
`(dV)/(dr) = E = (9xx10^(9)xx78xx1.6xx10^(-19))/((10^(-12))^(2))`
`= 1.123xx10^(17) Vm^(-1)`
At `r = 5xx10^(-15) m`,
`(dV)/(dr) = E = (q)/(4pi in_(0) r^(2)) = (9xx10^(9)xx78xx1.6xx10^(-19))/((5xx10^(-15))^(2))`
`= 4.5xx10^(21) Vm^(-1)`