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Question: Find the current produced at room temperature in a pure germanium plate of area \(2 \times {10^{ - 4...

Find the current produced at room temperature in a pure germanium plate of area 2×10−4m22 \times {10^{ - 4}}{m^2} and of thickness 1.2×10−3m1.2 \times {10^{ - 3}}m when a potential of 5V5V is applied across the faces. Concentration of carriers in germanium at room temperature is 1.6×1061.6 \times {10^6} per cubic metre. The mobilities of electrons and holes are 0.4m2V−1s−10.4{m^2}{V^{ - 1}}{s^{ - 1}} and 0.2m2V−1s−10.2{m^2}{V^{ - 1}}{s^{ - 1}} respectively. The heat energy generated in the plate in 100 second is:
A) 2.4×10−11J2.4 \times {10^{ - 11}}J
B) 3.4×10−11J3.4 \times {10^{ - 11}}J
C) 5.4×10−11J5.4 \times {10^{ - 11}}J
D) 6.4×10−11J6.4 \times {10^{ - 11}}J

Explanation

Solution

In order to get the solution of the given question, first of all we need to find the value of current. Since, charge carriers are given in the question so we need to find the conductivity as well, then with the help of it, current can easily be found. After that we need to apply the Heat energy formula to get the required solution of the given question.

Complete step by step solution:
The area of the germanium plate is given, a=2×10−4m2a = 2 \times {10^{ - 4}}{m^2}
And the thickness of germanium plate is given as, t=1.2×10−3mt = 1.2 \times {10^{ - 3}}m
Charge carriers is given by, n=1.6×106m−3n = 1.6 \times {10^6}{m^{ - 3}}
Mobility of electrons is given by, μe=0.4m2V−1s−1{\mu _e} = 0.4{m^2}{V^{ - 1}}{s^{ - 1}}
Mobility of holes is given by, μh=0.2m2V−1s−1{\mu _h} = 0.2{m^2}{V^{ - 1}}{s^{ - 1}}
Potential difference is given by, V=5VV = 5V
We need to find the conductivity.
σ=ne(μe+μh)\sigma = ne({\mu _e} + {\mu _h})
⇒σ=1.6×106×1.6×10−19(0.4+0.2)\Rightarrow \sigma = 1.6 \times {10^6} \times 1.6 \times {10^{ - 19}}(0.4 + 0.2)
∴σ=1.53×10−13Sm−1\therefore \sigma = 1.53 \times {10^{ - 13}}S{m^{ - 1}}
Now, we need to find the value of the current.
I=JAI = JA
⇒(σE)A=σ(Vt)A\Rightarrow (\sigma E)A = \sigma \left( {\dfrac{V}{t}} \right)A
⇒I=(1.53×10−13)(51.2×10−3)(2×10−4)\Rightarrow I = (1.53 \times {10^{ - 13}})\left( {\dfrac{5}{{1.2 \times {{10}^{ - 3}}}}} \right)(2 \times {10^{ - 4}})
∴I=1.28×10−13A\therefore I = 1.28 \times {10^{ - 13}}A
As, we know that heat energy of a body is given by, H=VItH = VIt
Now, let us put the values in the above equation.
On, putting the values in the above equation, we get
⇒H=5×1.28×10−13×100\Rightarrow H = 5 \times 1.28 \times {10^{ - 13}} \times 100
∴H=6.4×10−11J\therefore H = 6.4 \times {10^{ - 11}}J

Hence, option (D), i.e. 6.4×10−11J6.4 \times {10^{ - 11}}J is the correct solution for the given question.

Note: In the given question, we need to take care of all the relation between, current, voltage, charge density, current density etc. We need to be clear with the formula used. Here, ‘J’ is the current density. Also, we know that the heat produced in the material depends on the current supply. And current depends on the flow of charge per unit time. Conductivity is the measure of the ease at which an electric charge or heat can pass through a material.