Class 12th

13.5 Mass of the copper coin, m' = 3.0 g

Atomic mass of ${}_{29}{}^{63}Cu$ , m = 62.92960 u

The total number of ${}_{29}{}^{63}Cu$ atoms in the coin, N = $\frac{N_A*m\text{'}}{Massnumber}$ , where

$N_A$ = Avogadro's number = 6.023 $*{10}^{23}$ atoms / g

Mass number = 63 g

Therefore, N = $\frac{6.023\mathrm{}*{10}^{23}*3}{63}$ = 2.868 $*{10}^{22}$ atoms

${}_{29}{}^{63}Cu$ has 29 protons and (63 – 29) 34 neutrons

Hence the mass defect of the nucleus Δm = 29 $*m_p$ + 34 $*m_n$ - $m$

Mass of a proton, $m_p$ = 1.007825 u

Mass of a neutron, $m_n$ = 1.008665 u

Δm = 29 $*1.007825\mathrm{}$ + 34 $*1.008665\mathrm{}$ - 62.92960

Δm = 0.591935 u

Mass defect of all the atoms present in the coin, Δm = 0.591935 $*N$

= 0.591935 $*$ 2.868 $*{10}^{22}$ = 1.69766958 $*{10}^{22}$ u

But 1 u = 931.5 MeV/ $c^2$

Δm = 1.581 $*{10}^{25}$ MeV/ $c^2$

The binding energy of the nucleus, $E_b$ = Δm $c^2$ , where c = speed of light

$E_b=$ (1.581 $*{10}^{25}$ / $c^2$ ) $*$ $c^2$ = 1.581 $*{10}^{25}$ MeV

1 MeV = 1.6 $*{10}^{-13}$ J

Hence, $E_b=$ 2.530 $*{10}^{12}$ J

This much of energy required to separate all the neutrons and protons from the given coin.

14.15 A acts as two inputs of the NOR gate and Y is the output. As shown in the following figure. Hence the output of the circuit is $\stackrel{?}{A+A}$ = $\stackrel{?}{A}$

The truth table for the same is given as:

This is the truth table of a NOT gate. Hence, this circuit functions as a NOT gate.

A and B are the inputs and Y is the output of the given circuit. By using the result obtained in solution (a), we can infer that the outputs of the first two NOR gates are $\stackrel{?}{A}$ and $\stackrel{?}{B}$  , as shown in the following figure

Above is given the inputs for the last NOR gate.

Hence, the output for the circuit can be written as:

Y = $\stackrel{?}{A+B}$ = $\stackrel{?}{\stackrel{?}{A}.\stackrel{?}{B}}$   = A.B

The truth table for the same can be written as:

This is the truth table of an AND gate. Hence, this circuit functions as an AND gate.

14.10 In a p-n junction diode, the expression for current is given as:

$\mathit{I}=\mathit{}{\mathit{I}}_{0\mathit{}}\mathit{e}\mathit{x}\mathit{p}\left(\frac{\mathit{e}\mathit{V}}{2{\mathit{k}}_{\mathit{B}}\mathit{T}}-1\right)$

Where, $I_0$ = Reverse saturation current = 5 $*{10}^{-12}$ A

T = Absolute temperature = 300 K

${\mathit{k}}_{\mathit{B}}$ = Boltzmann constant = 8.6  $*{10}^{-5}$  eV/K = 8.6 $*{10}^{-5}*1.6*{10}^{-19}$ J/K = 1.376 $$ $*{10}^{-23}J/K$

V = voltage across the diode

e = charge of an electron = 1.6 $*{10}^{-19}$ C

Forward voltage, V = 0.6V

Current, $I=5*{10}^{-12}*exp\left(\frac{1.6*{10}^{-19}*0.6}{1.376*{10}^{-23}*300}-1\right)$ = $5*{10}^{-12}*\mathrm{e}\mathrm{x}\mathrm{p}?\left(22.256\right)$

= 0.02315 A

For forward voltage, V = 0.7V, we can write

Current, $I\text{'}=5*{10}^{-12}*exp\left(\frac{1.6*{10}^{-19}*0.7}{1.376*{10}^{-23}*300}-1\right)$ = $5*{10}^{-12}*\mathrm{e}\mathrm{x}\mathrm{p}?\left(26.132\right)$

= 1.117 A

Hence increase in the current, ΔI = I'-I = 1.117 – 0.02315 = 1.0934 A

Dynamic resistance =  $\frac{Changeinvoltage}{ChangeinCurrent}$ = $\frac{0.7-0.6}{1.0934}=0.091\Omega$

If the reverse bias voltage is changed from 1 V to 2 V, the current will remain same as $I_0$ will be equal in both cases. Therefore the dynamic resistance in the reverse bias will be $\frac{1}{0}=\infty$

14.9 Energy gap of the given intrinsic semiconductor,  $E_g$  = 1.2 eV

The temperature dependence of the intrinsic carrier-concentration is written as:

$n_i=n_{0}exp\left(-\frac{E_g}{2k_{B}T}\right)$

Where,  $k_B$ = Boltzmann constant = 8.62 $*{10}^{-5}$
 eV/K

T = Temperature

$n_0$ = constant

Initial temperature, $T_1$
 = 300 K

The intrinsic carrier-concentration at this temperature can be written as

$n_{i1}=n_{0}exp\left(-\frac{E_g}{2k_B*300}\right)$ ……(1)

Final temperature,  $T_2$ = 600 K

The intrinsic carrier-concentration at this temperature can be written as

$n_{i2}=n_{0}exp\left(-\frac{E_g}{2k_B*600}\right)$ ……(2)

The ratio between conductivity at 600K and at 300 K is equal to the ratio between the respective intrinsic carrier concentration at these temperatures.

Therefore,

$\frac{n_{i2}}{n_{i1}}$ =  $\frac{n_{0}exp\left(-\frac{E_g}{2k_B*600}\right)}{n_{0}exp\left(-\frac{E_g}{2k_B*300}\right)}$

= exp $\frac{E_g}{2k_B}\left[\frac{1}{300}-\frac{1}{600}\right]$

= exp $\frac{E_g}{2k_B}\left[\frac{1}{300}-\frac{1}{600}\right]$

= exp $\frac{1.2}{2*8.62*{10}^{-5}}\left[\frac{2-1}{600}\right]$  = exp(11.6) = 109 $\frac{{}_{}}{}$ $*{10}^5$

Therefore, the ratio between two conductivities is 1.09 $*{10}^5$

$\mathit{I}=\mathit{}{\mathit{I}}_{0\mathit{}}\mathit{e}\mathit{x}\mathit{p}\left(\frac{\mathit{e}\mathit{V}}{2{\mathit{k}}_{\mathit{B}}\mathit{T}}-1\right)$