Answer:

Explanation:
Due to Coulomb´s law electric force can be described by the formula
, where K is the Coulomb´s constant (
),
= Charge 1 (Na+ in this case),
is the charge 2 (Cl-) and r is the distance between both charges.
Work made by a force is W=F.d and total work produced is the change in energy between final and initial state. this is
.
so we have ![W=W_{f} -W_{i} =(K\frac{q_{(Na+)}q_{(Cl-)}rf}{r_{f} ^{2}})-(K\frac{q_{(Na+)}q_{(Cl-)}ri}{r_{i} ^{2}})=Kq_{(Na+)}q_{(Cl-)[\frac{1}{{r_{f}}} -\frac{1}{{r_{i}}}]](https://tex.z-dn.net/?f=W%3DW_%7Bf%7D%20-W_%7Bi%7D%20%3D%28K%5Cfrac%7Bq_%7B%28Na%2B%29%7Dq_%7B%28Cl-%29%7Drf%7D%7Br_%7Bf%7D%20%5E%7B2%7D%7D%29-%28K%5Cfrac%7Bq_%7B%28Na%2B%29%7Dq_%7B%28Cl-%29%7Dri%7D%7Br_%7Bi%7D%20%5E%7B2%7D%7D%29%3DKq_%7B%28Na%2B%29%7Dq_%7B%28Cl-%29%5B%5Cfrac%7B1%7D%7B%7Br_%7Bf%7D%7D%7D%20-%5Cfrac%7B1%7D%7B%7Br_%7Bi%7D%7D%7D%5D)
Given that ri= 1.1nm=
and rf= infinite distance
![W=(9x10^{9})(1.6x10^{-19})(-1.6x10^{-19})[\frac{1}{\alpha }-\frac{1}{(1.1x10^{-9})}]=2.1x10^{-19}J](https://tex.z-dn.net/?f=W%3D%289x10%5E%7B9%7D%29%281.6x10%5E%7B-19%7D%29%28-1.6x10%5E%7B-19%7D%29%5B%5Cfrac%7B1%7D%7B%5Calpha%20%7D-%5Cfrac%7B1%7D%7B%281.1x10%5E%7B-9%7D%29%7D%5D%3D2.1x10%5E%7B-19%7DJ)
Answer: Wavenumber of the radiation emitted is 
Explanation:
The relationship between wavelength and energy of the wave follows the equation:

where,
E = energy of the radiation = 
h = Planck's constant = 
c = speed of light = 
= wavelength of radiation = ?
Putting values in above equation, we get:


Thus wavenumber of the radiation emitted is 
Answer:
Explanation:
"Three hundred ten million, seven hundred sixty-three thousand, one hundred thirty-six".
An organic compound that contains a carbonyl group with a hydroxyl group attached to it is an example of a (d) carboxylic acid.