
Find the ratio of the conduction electrons to the total number of atoms in the given conductor.
(A). Almost 1:4
(B). Almost 1:2
(C). Almost 2:4
(D). Almost 1:1
Answer
572.1k+ views
Hint: The formula used in this question is of current density. \[J = \dfrac{i}{A}\] Current \[i\]be spread uniformly over the area of cross-section \[A\]of a conductor. Along with it we should know about conductors, drift velocity and mobility of electrons. We will be learning each concept with all the related information followed by the calculation part which will help us in approaching our solution.
Complete step-by-step answer:
Conductor – Materials in which a charge moiety can make movement without restrictions. Inside conductor electrostatic field is zero. In conductors, surface charge density always varies with position.
Drift Velocity – It is coined as the average velocity through which the free electrons in a conductor get moved towards the positive end of the conductor below the influence of an electric field that persists across the conductor. It is denoted by\[{V_d}\]
\[F\]=\[ - e\]\[E\]
Mobility – It is the magnitude of drift velocity of charge divided by unit electric field applied. It is expressed as,
\[\mu = \dfrac{{qE\tau /m}}{E} = \dfrac{{q\tau }}{m}\]
Where, \[\tau \] is the average relaxation time of the charge while drifting towards the opposite electrode and \[m\]is the mass of the charged particle.
Mobility of electrons, \[{\mu _e} = \dfrac{{e{\tau _e}}}{{{m_e}}}\]
Using, \[j = nq{v_d} \Rightarrow {v_d} = \dfrac{j}{{nq}}........1\]
Now, \[F = q{E_x} = q{v_d}B\]
Thus, \[{n_e} = \dfrac{{jB}}{{eE}}\]
Given; \[E = 500\mu V/mj = 200 \times {10^4}A/{m^2}B = 1T\]
Thus, \[{n_e} = 2.5 \times {10^{22}}/cc\]
Now, Atomic weight of sodium is 23 and density\[ \approx 1 \Rightarrow \]molar volume\[ = 23cc\], thus number of atom per unit volume
\[{n_a}\]\[ = \dfrac{{6 \times {{10}^{23}}}}{{23}} = 2.6 \times {10^{22}}\]
Hence, \[{n_e}:{n_a} \approx 1:1\]
Hence, the correct answer is option (D) Almost $1:1$
Note: Current Density is coined as the ratio of the current at that point in the conductor to the area of cross-section of a conductor at that point. It is the characteristic property of a point inside the conductor. It is a vector quantity.
Complete step-by-step answer:
Conductor – Materials in which a charge moiety can make movement without restrictions. Inside conductor electrostatic field is zero. In conductors, surface charge density always varies with position.
Drift Velocity – It is coined as the average velocity through which the free electrons in a conductor get moved towards the positive end of the conductor below the influence of an electric field that persists across the conductor. It is denoted by\[{V_d}\]
\[F\]=\[ - e\]\[E\]
Mobility – It is the magnitude of drift velocity of charge divided by unit electric field applied. It is expressed as,
\[\mu = \dfrac{{qE\tau /m}}{E} = \dfrac{{q\tau }}{m}\]
Where, \[\tau \] is the average relaxation time of the charge while drifting towards the opposite electrode and \[m\]is the mass of the charged particle.
Mobility of electrons, \[{\mu _e} = \dfrac{{e{\tau _e}}}{{{m_e}}}\]
Using, \[j = nq{v_d} \Rightarrow {v_d} = \dfrac{j}{{nq}}........1\]
Now, \[F = q{E_x} = q{v_d}B\]
Thus, \[{n_e} = \dfrac{{jB}}{{eE}}\]
Given; \[E = 500\mu V/mj = 200 \times {10^4}A/{m^2}B = 1T\]
Thus, \[{n_e} = 2.5 \times {10^{22}}/cc\]
Now, Atomic weight of sodium is 23 and density\[ \approx 1 \Rightarrow \]molar volume\[ = 23cc\], thus number of atom per unit volume
\[{n_a}\]\[ = \dfrac{{6 \times {{10}^{23}}}}{{23}} = 2.6 \times {10^{22}}\]
Hence, \[{n_e}:{n_a} \approx 1:1\]
Hence, the correct answer is option (D) Almost $1:1$
Note: Current Density is coined as the ratio of the current at that point in the conductor to the area of cross-section of a conductor at that point. It is the characteristic property of a point inside the conductor. It is a vector quantity.
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