Question

a uranium atom emits Alpha particle if at any instant Alpha particle is at a distance of 9× 10^-15 m from the centre of the disintegrated atom then what will be the force acting on the Alpha particle​

Answer 1

The force on the alpha particle is 512 N

Explanation:

We are given that:

Parent nuclei is U 92 - 238

Emitted particle = He 2- 4

Remaining daughter nuclei = X 92 − 2^238 − 4 = X 90^234

Charge of daughter nuclei = 90 × charge on proton

charge on alpha particle = 2 × charge on proton

Distance between them r = 9 × 10^−15 m

Electric force of repulsion between them

F = k q2 × q1  / r^2

F = 9 × 10^9 × 90 × 2 × (1.6 × 10^−19)^2 / (9 × 10^−15)^2

F = 51.2 × 10 = 512 N

Hence the force on the alpha particle is 512 N