
The maximum Coulomb force that can act on the electron due to the nucleus in a hydrogen atom will be:
(A) \[0.82 \times {10^{ - 8}}\]
(B) \[0.082 \times {10^{ - 8}}\]
(C) \[8.2 \times {10^{ - 8}}\]
(D) \[82 \times {10^{ - 8}}\]
Answer
232.8k+ views
Hint Coulomb force is the force due to electric charges. Charge of the electron and nucleus with distance separating them is substituted in \[F = \dfrac{{{q_1}{q_2}}}{{4\pi {\smallint _0}{r^2}}}\]. Maximum of this force means minimum radius. And the minimum radius is when the electron is in the ground level.
Complete step-by-step answer
Using the Coulomb force law
\[F = \dfrac{{{q_1}{q_2}}}{{4\pi {\smallint _0}{r^2}}}\]
$q_1$= charge of nucleus\[ = 1.6\times{10^{ - 19}}\] C
Charge of the nucleus will be equal to the charge of protons inside the nucleus as the neutrons do not contribute towards the charge. For a hydrogen atom, there is only 1 proton in the nucleus.
Higher force means: more charge and less radius. But in an atom we usually only have definite charge. But the smallest radius is obtained when the electron is in the smallest orbit. In our case this is the first orbit
$q_2$ = charge of electron\[ = 1.6 \times {10^{ - 19}}\]C
R= distance between the nucleus and the electron\[ = 0.53 \times {10^{ - 10}}\]m
E0= permittivity of free space\[\; = 8.85418782{\text{ }} \times {\text{ }}{10^{ - 12}}\;{m^{ - 3}}\;k{g^{ - 1}}\;{s^4}\;{A^2}\]
\[F = \dfrac{{{{(1.6 \times {{10}^{ - 19}})}^2}}}{{4\pi {\smallint _0}0.53 \times {{10}^{ - 10}}}}\]
\[F = \dfrac{{{{(1.6 \times {{10}^{ - 19}})}^2} \times 9 \times {{10}^9}}}{{0.53 \times {{10}^{ - 10}}}}\]
\[F = 8.202 \times {10^{ - 8}}\] N
Therefore, the correct answer is C
Note The electron will experience the same force as long as it stays in the orbit of ground level. Electrons in higher energy levels will be further away from the nucleus; hence the Coulomb force is doing to drop. Also, when the electron is moving in a circular orbit around the nucleus, it will also experience centripetal force, which will be equal and opposite to the force of attraction between the electron and nucleus.
Complete step-by-step answer
Using the Coulomb force law
\[F = \dfrac{{{q_1}{q_2}}}{{4\pi {\smallint _0}{r^2}}}\]
$q_1$= charge of nucleus\[ = 1.6\times{10^{ - 19}}\] C
Charge of the nucleus will be equal to the charge of protons inside the nucleus as the neutrons do not contribute towards the charge. For a hydrogen atom, there is only 1 proton in the nucleus.
Higher force means: more charge and less radius. But in an atom we usually only have definite charge. But the smallest radius is obtained when the electron is in the smallest orbit. In our case this is the first orbit
$q_2$ = charge of electron\[ = 1.6 \times {10^{ - 19}}\]C
R= distance between the nucleus and the electron\[ = 0.53 \times {10^{ - 10}}\]m
E0= permittivity of free space\[\; = 8.85418782{\text{ }} \times {\text{ }}{10^{ - 12}}\;{m^{ - 3}}\;k{g^{ - 1}}\;{s^4}\;{A^2}\]
\[F = \dfrac{{{{(1.6 \times {{10}^{ - 19}})}^2}}}{{4\pi {\smallint _0}0.53 \times {{10}^{ - 10}}}}\]
\[F = \dfrac{{{{(1.6 \times {{10}^{ - 19}})}^2} \times 9 \times {{10}^9}}}{{0.53 \times {{10}^{ - 10}}}}\]
\[F = 8.202 \times {10^{ - 8}}\] N
Therefore, the correct answer is C
Note The electron will experience the same force as long as it stays in the orbit of ground level. Electrons in higher energy levels will be further away from the nucleus; hence the Coulomb force is doing to drop. Also, when the electron is moving in a circular orbit around the nucleus, it will also experience centripetal force, which will be equal and opposite to the force of attraction between the electron and nucleus.
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