
An oil drop has \[6.39 \times {10^{ - 19}}C\;\] charge. What will be the number of electrons in this drop?
A.2
B.4
C.8
D.16
Answer
559.5k+ views
Hint:In this question we will use the relation between charge, number of electrons and the charge on one electron i.e. $Q = ne$, here Q is charge, n is number of electrons and e is the charge on one electron.
Complete step by step answer:
Atom of an element is its basic component of an element and has electrons, protons and neutrons in its structure. The electrons are present in the shells surrounding the nucleus and protons and neutrons are present in the nucleus of the atom.
Electrons: They are negatively charged ions present in the shells of an atom.
Mass of an electron is fixed which is$9.01 \times {10^{ - 31}}{\text{kilogram}}$.
Charge of electron: $ - 1.602 \times {10^{ - 19}}$
Protons: They are positively charged ions present in the nucleus of the atom.
Charge of proton: $1.602 \times {10^{ - 19}}$
According to the question the charge on oil drop =\[6.39 \times {10^{ - 19}}C\;\]
we know that charge on one electron is equal to $1.602 \times {10^{ - 19}}$
To calculate the number of electrons given, the formula used will be;
$Q = ne..........(1)$, here $Q$ stands for charge, n is the number of electrons and e is the charge on one electron.
Substituting the values in equation (1).
$Q = ne$
$n = {\text{ }}\dfrac{Q}{e}$
ANd hence on doing the simplification,we have
n$ = {\text{ }}\dfrac{{6.39 \times {{10}^{ - 19}}C}}{{1.602 \times {{10}^{ - 19}}}}\;$
n $ = 4$
So the total number of electrons present in oil drop is = $4$
Thus the correct option is B.
Note:
Neutrons: They are the neutral species present along the protons in the nucleus of an atom.
The charge of nucleus is due to presence of protons and contribution of neutrons is nil to the charge of nucleus, but the mass of both neutron and proton is quite similar which is $1.67 \times {10^{ - 24}}gram$.
The mass of neutron combined with the mass of protons, gives nucleus its density and hardness and the mass of electron is negligible in comparison to nucleus mass
Complete step by step answer:
Atom of an element is its basic component of an element and has electrons, protons and neutrons in its structure. The electrons are present in the shells surrounding the nucleus and protons and neutrons are present in the nucleus of the atom.
Electrons: They are negatively charged ions present in the shells of an atom.
Mass of an electron is fixed which is$9.01 \times {10^{ - 31}}{\text{kilogram}}$.
Charge of electron: $ - 1.602 \times {10^{ - 19}}$
Protons: They are positively charged ions present in the nucleus of the atom.
Charge of proton: $1.602 \times {10^{ - 19}}$
According to the question the charge on oil drop =\[6.39 \times {10^{ - 19}}C\;\]
we know that charge on one electron is equal to $1.602 \times {10^{ - 19}}$
To calculate the number of electrons given, the formula used will be;
$Q = ne..........(1)$, here $Q$ stands for charge, n is the number of electrons and e is the charge on one electron.
Substituting the values in equation (1).
$Q = ne$
$n = {\text{ }}\dfrac{Q}{e}$
ANd hence on doing the simplification,we have
n$ = {\text{ }}\dfrac{{6.39 \times {{10}^{ - 19}}C}}{{1.602 \times {{10}^{ - 19}}}}\;$
n $ = 4$
So the total number of electrons present in oil drop is = $4$
Thus the correct option is B.
Note:
Neutrons: They are the neutral species present along the protons in the nucleus of an atom.
The charge of nucleus is due to presence of protons and contribution of neutrons is nil to the charge of nucleus, but the mass of both neutron and proton is quite similar which is $1.67 \times {10^{ - 24}}gram$.
The mass of neutron combined with the mass of protons, gives nucleus its density and hardness and the mass of electron is negligible in comparison to nucleus mass
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