Every atom makes one free electron in copper. If $1.1$ $ampere$ current is flowing in the wire of copper having $1\, mm$ diameter, then the drift velocity (approx.) will be (Density of copper $ = 9 \times {10^3}\,kg\,{m^{ - 3}}$ and atomic weight = $63$)
Diffcult
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(b) Density of $Cu = 9 \times {10^3}\,kg/{m^3}$ (mass of $1$ $m^3$ of $Cu$)
( $6.0 \times 10^{23}$ atoms has a mass = $63 \times 10^{-3}\, kg$
Number of electrons per $m^3$ are
$ = \frac{{6.0 \times {{10}^{23}}}}{{63 \times {{10}^{ - 3}}}} \times 9 \times {10^3} = 8.5 \times {10^{28}}$
Now drift velocity $ = {v_d} = \frac{i}{{neA}}$
$ = \frac{{1.1}}{{8.5 \times {{10}^{28}} \times 1.6 \times {{10}^{ - 19}} \times \pi \times {{(0.5 \times {{10}^{ - 3}})}^2}}}$
$ = 0.1 \times {10^{ - 3}}\,m\,/sec$
( $6.0 \times 10^{23}$ atoms has a mass = $63 \times 10^{-3}\, kg$
Number of electrons per $m^3$ are
$ = \frac{{6.0 \times {{10}^{23}}}}{{63 \times {{10}^{ - 3}}}} \times 9 \times {10^3} = 8.5 \times {10^{28}}$
Now drift velocity $ = {v_d} = \frac{i}{{neA}}$
$ = \frac{{1.1}}{{8.5 \times {{10}^{28}} \times 1.6 \times {{10}^{ - 19}} \times \pi \times {{(0.5 \times {{10}^{ - 3}})}^2}}}$
$ = 0.1 \times {10^{ - 3}}\,m\,/sec$
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