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
The Rorschach inkblots and the TAT (Thematic Appreciation Test) both rely on providing the subject with ambiguous visual stimuli and assessing the subject's state of mind using the subject's interpretation of the stimuli.
Both use cards, although not all of the cards are used in the TAT. Moreover, the TAT cards contain sketches, while the Rorschach inkblots contain patterns of ink.
Potential energy is highest at the top of the loop, and kinetic energy is highest at the bottom of the loop.
It takes more energy to remove the second electron from a lithium atom than it does to remove the fourth electron from a carbon atom because its inner core e, not valence e. C's 4th removed e is still a valence e. And also <span>because more nuclear charge acting on the second electron, it is more close to the nucleus, thus the the protons attract it more than the 4th electron.</span>
