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

Answer:

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

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 = $\frac{1}{4 \pi \epsilon _o} \frac{q1q2}{r^2}$    .............1

put here value we get

electrical force = $9*10^9 \frac{92*(1.6*10^{-19})^2}{(1.75*10^{-10})^2}$

electrical force =  6.921 × $10^{-7}$ N

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

Answer 2

Answer:

6.92 x 10^-7 N

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

$F = \frac{KQq}{d^{2}}$

$F = \frac{9\times 10^{9}\times 147.2 \times 10^{-19}\times 1.6\times 10^{-19}}{\left ( 175\times 10^{-12} \right )^{2}}$

F = 0.0692 x 10^-5 N

F = 6.92 x 10^-7 N

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