$(a)$ Consider circuit in figure. How much energy is absorbed by electrons from the initial state of no current (ignore thermal motion) to the state of drift velocity ?

$(b)$ Electrons give up energy at the rate of $R{I^2}\;$ per second to the thermal energy. What time scale would number associate with energy in problem $(a)$ ? $n = no$ of electron/volume $ = {10^{29}}{m^{ - 3}}$, length of circuit $= 10$ $cm$, cross-section $=$ $A = $ ${\left( {1\,mm} \right)^2}$

$(a)$ Current in the circuit $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6}{6}=1 \mathrm{~A}$

Number density of electron $n=\frac{10^{29}}{\mathrm{~m}^{3}}$

Length of wire of circuit $l=10 \mathrm{~cm}=0.1 \mathrm{~m}$

Area of cross section $A=(1 \mathrm{~mm})^{2}$

$\Rightarrow$ Current due to electron moving with drift velocity,

$\mathrm{I}=n \mathrm{~A} v_{d} e$

$\therefore v_{d} =\frac{\mathrm{I}}{n \mathrm{~A} e}$

$=\frac{1}{10^{29} \times\left(10^{-3}\right)^{2} \times 1.6 \times 10^{-19}}$4

$=\frac{10^{-5}}{16} \frac{\mathrm{m}}{\mathrm{s}}=0.0625 \times 10^{-5} \frac{\mathrm{m}}{\mathrm{s}}$

$=6.25 \times 10^{-3} \frac{\mathrm{m}}{\mathrm{s}}$

$\Rightarrow$ Energy absorbed in form of $K.\;E$.,

$\mathrm{K.E} .= \frac{1}{2} m_{e} v_{d}^{2} \times n \mathrm{Al}$

$=\frac{1}{2} \times 9.1 \times 10^{-31} \times\left(6.25 \times 10^{-3}\right)^{2}$

$\times 10^{29} \times 10^{-6} \times 10^{-1}$

$=177.7 \times 10^{-19}$

$\therefore \quad \text { K.E. } \approx 1.78 \times 10^{-17} \mathrm{~J}$

$(b)$ Power dissipated,

$\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$

$\frac{\mathrm{E}}{t}=(1)^{2} \times 6$

$\therefore \frac{\mathrm{E}}{t}=6 \mathrm{~W}$

$\therefore t=\frac{\mathrm{E}}{6}=\frac{2 \times 10^{-17}}{6} \quad[\because \mathrm{K.\textrm{E } . = \mathrm { E } ]}$

$\therefore t =\frac{1}{3} \times 10^{-17}$

$\therefore t=0.33 \times 10^{-17} \mathrm{~s}$

$\therefore T \approx 3.3 \times 10^{-18} \mathrm{~s}$