Question

What electrical force dies a Uranium nucleus exert on one of its inner electrons, located at a distance of 175 picometers (= 1.75 x 10^-10 m)?

a) 1.2 x 10^-16 N
b) 7.2 x 10^11 N
c) 7.5 x 10^-9 N
d) 7.0 x 10^-7 N
e) 1.8 x 10^-6 N

Answer 1

6.92 x 10^-7 N

Step-by-step explanation:

distance, d = 175 x 10^-12 m

charge on electron, q = 1.6 x 10^-19 C

charge on nucleus, Q = 92 x 1.6 x 10^-19 C = 147.2 x 10^-19 C

The force between the electron and the nucleus is given by

`F = (KQq)/(d^(2))`

`F = (9* 10^(9)* 147.2 * 10^(-19)* 1.6* 10^(-19))/(\left ( 175* 10^(-12) \right )^(2))`

F = 0.0692 x 10^-5 N

F = 6.92 x 10^-7 N

Thus, the force is 6.92 x 10^-7 N.

Answer 2

correct option is d) 7.0 x 10^-7 N

Step-by-step explanation:

given data

distance = 175 picometers = 1.75 ×
`10^(-10)` m

to find out

electrical force

solution

we know atomic no of uranium is 92

and charge on electron is = 1.6 ×
`10^(-19)` C

and electrical force is express as

electrical force =
`(1)/(4 \pi \epsilon _o) (q1q2)/(r^2)` .............1

put here value we get

electrical force =
`9*10^9 (92*(1.6*10^(-19))^2)/((1.75*10^(-10))^2)`

electrical force = 6.921 ×
`10^(-7)` N

so correct option is d) 7.0 x 10^-7 N