Question

The amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is__________. Let the radius (r) of atom is 52 pm a) Zero b) infinity c) Fr d) 2Fr

Answer 1

Answer:

d) 2Fr

Explanation:

We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell  -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².

We now evaluate the integral from r = +r to -r

W = ∫Fdr

= ∫(-e²/4πε₀r²)dr

= -∫e²dr/4πε₀r²

= -e²/4πε₀∫dr/r²

= -e²/4πε₀ × -[1/r] from r = +r to -r

W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.

Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.

So W = -2e²/4πε₀r = 2Fr.

So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr