91影视

The given data can be listed below as:

• The applied potential difference is V = 120 V.
• The dissipated power is, P = 500W.

In this problem, by using equation (i), we can find the resistance during operation. Using equation (iii) in the formula for the rate of electron transport, we can find the rate of electron flow through any cross-section of the heater element.

Formulae:

The electric power generated during the energy transfer is,

$P=\frac{V^2}{R}$ 鈥� (i)

The current flowing due to energy transfer is,

$i=\frac{P}{V}$ 鈥� (ii)

The rate of electron transport due to the transfer is,

$\mathrm{Electron}\mathrm{transport}\mathrm{rate}=\frac{i}{e}$ 鈥� (iii)

Using the given data in equation (i), we can get the value of the resistance during the operations as follows:

$R=\frac{V^2}{P}$

Substitute the values in the above equation.

$$\begin{gathered} R=\frac{{(120\mathrm{V})}^2}{500\mathrm{W}} \\ =28.8\mathrm{惟} \end{gathered}$$

Hence, the value of the resistance is $28.8\mathrm{惟}$.

Using equation (ii) in equation (iii) with the given values, we can get the value of the electron flow rate through any cross-section of the heater element as follows:

$\mathrm{Electron}\mathrm{transport}\mathrm{rate}=\frac{P}{eV}$

Here, e is the electric charge whose value is$e=1.60脳{10}^{-19}\mathrm{C}$.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{Electron}\mathrm{transport}\mathrm{rate}=\frac{500\mathrm{W}}{1.60脳{10}^{-19}\mathrm{C}脳120\mathrm{V}} \\ =2.60脳{10}^{19}/\mathrm{s} \end{gathered}$$

Hence, the value of the electron flow rate is $2.60脳{10}^{19}/\mathrm{s}$.