Question
Question: In \[1g\] of solid there are \(5 \times {10^{21}}\) atoms. If one electron is removed from the energ...
In 1g of solid there are 5×1021 atoms. If one electron is removed from the energy of 0.01% atoms of the solid then the charge gained by the solid is, (electronic charge is 1.6×1019C)
(A) +0.08C
(B) +0.8C
(C) −0.08C
(D) −0.8C
Solution
Charge is a quantised quantity and charge of a single electron is known as a quantum of charge. This theory can be used in order to solve the above problem. We will also have to keep in mind the sign of the charge gained which is determined by the excess or deficiency of the electrons.
Complete step by step answer:
Let’s start with the problem itself. It says that the solid contains 5×1021 atoms and we removed 0.01% atoms from the primary given mass.
We can conclude at this point that as charge is being removed from the solid it will result in a positive sign of the net charge gained by the solid. But let’s not jump into that now.
Let’s first determine the number of electrons removed from the solid in the first place.
If n is the number of electrons removed from the solid then
n=1000.01×5×1021
This is the 0.01% of all the atoms in the solid.
⇒n=5×1017
So as we have removed n number of electrons then the amount of charge removed is given by,
q=ne
Where e is the charge of a single electron.
Putting the respective values in the above equation we get,
⇒q=5×1017×(1.6×10−19)
⇒q=−0.08C
The minus sign here implies that a charge equivalent to −0.08C is removed from the solid.
This draws us to the conclusion that the solid gains a charge of +0.08C due to removal of q amount of charge.
So the correct answer is option A.
Note:
The formula q=ne is known as the principle of quantization of electric charge, which means that any amount of charge in the universe can be represented as an integral multiple of charge of a single electron. And the charge of a single electron is known as a quantum of charge.