Class 12th

11.4 Wavelength of the monochromatic light, $\lambda =632.8nm$ = 632.8 $*{10}^{-9}$ m

Power emitted by laser, P = 9.42 mW = 9.42 $*{10}^{-3}$ W

Planck's constant, h = 6.626 $*{10}^{-34}$ Js

Speed of light, c = 3 $*{10}^8$ m/s

Mass of hydrogen atom, m = 1.66 $*{10}^{-27}$ kg

The energy of each photon is given as, $E$ = $\frac{hc}{\lambda }$ = $\frac{6.626\mathrm{}*{10}^{-34}*3\mathrm{}*{10}^8}{632.8*{10}^{-9}}$ J = 3.141 $*{10}^{-19}$ J

The momentum of each photon is given by $p$ = $\frac{h}{\lambda }$ = $\frac{6.626\mathrm{}*{10}^{-34}}{632.8*{10}^{-9}}$ = 1.047 $*{10}^{-27}$ kg-m/s

Number of photons arriving per second at a target irradiated by the beam = n

Assume that the beam has a uniform cross-section that is less than the target area.

Hence, the equation for power can be written as, $P=nE$

n = $\frac{P}{E}$ = $\frac{9.42*{10}^{-3}}{3.141*{10}^{-19}}$ = 3.0 $*{10}^{16}$

Momentum of the hydrogen atom = momentum of the proton = 1.047 $*{10}^{-27}$ kg-m/s

The momentum of the hydrogen atom , $p=mv$ , where m = mass of hydrogen atom and v = velocity

Hence, v = $\frac{p}{m}$ = $\frac{1.047*{10}^{-27}}{1.66\mathrm{}*{10}^{-27}}$ = 0.631 m/s

11.2 Work function of cesium metal, ${\varnothing }_0$ = 2.14 eV

Frequency of light, $\nu$ = 6 $*{10}^{14}$ Hz

The maximum kinetic energy is given by the photoelectric effect, $K$ = $h\nu -{\varnothing }_0$ ,

Where $h$ = Planck's constant = 6.626 $*{10}^{-34}$ Js, 1 eV = 1.602 $*{10}^{-19}$ J

$K$ = $\frac{6.626*{10}^{-34}*6*{10}^{14}}{1.602*{10}^{-19}}-2.14$ = 0.345 eV

$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{x}\mathrm{i}\mathrm{m}\mathrm{u}\mathrm{m}\mathrm{}\mathrm{k}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{e}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{d}\mathrm{}\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}0.3416\mathrm{}\mathrm{e}\mathrm{V}$

For stopping potential $V_0$ , we can write the equation for kinetic energy as:

$K$ = $V_0$ , where e = charge of an electron = $1.6*{10}^{-19}$

or $V_0=\frac{K}{e}$ = $\frac{0.345*1.602*{10}^{-19}}{1.6*{10}^{-19}}$ = 0.345 V

Hence, the stopping potential is 0.345 V

Maximum speed of the emitted photoelectrons = v

Kinetic energy K = $\frac{1}{2}$ m $v^2$ , where m = mass of the electron = 9.1 $*{10}^{-31}$ kg

Hence, v = $\sqrt{\frac{2K}{m}}$ = $\sqrt{\frac{2*0.345*1.602*{10}^{-19}}{9.1\mathrm{}*{10}^{-31}}}$ = 348.5 $*{10}^3$ m/s = 346.8 km/s

Therefore, the maximum speed of the emitted photoelectrons is 346.8 km/s.

7.26 The rating of step-down transformer = 40000 V – 220 V

Hence, the input voltage,  $V_1$ = 40000 V

Output voltage,  $V_2$ = 220 V

Total electric power required, P = 800 kW = 800 $*{10}^3$ W

Source potential, V = 220 V

Voltage at which electric plant generates power, V' = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wires lines carrying power = 0.5 Ω /km

Total resistance of the wire, R = 2 $*15*0.5\Omega$ = 15 Ω

rms current in the wire lines is given as

I = $\frac{P}{V_1}$ = $\frac{800\mathrm{}*{10}^3}{40000}$ = 20 A

Line power loss = $I^{2}R$ = ${20}^2*$ 15 = 6 $*{10}^3$ W = 6 kW

Since the leakage power loss is negligible, the total power to be supplied by plant = 800 + 6 = 806 kW

Voltage drop in the transmission line = IR = 20 $*15$ = 300 V. Hence, the total voltage transmitted from plant = 300 + 40000 = 40300 V

Since the power generated is 440V, the transformer rating at the plant should be 440 V – 40300 V

Hence, power loss during transmission = $\frac{6}{806}$ $*100$ = 0.744 %

In the previous exercise, power loss during transmission = $\frac{600}{1400}$ $*100$ = 42.86 %

Since the power loss is less for a high voltage transmission, high voltage transmission is preferred.

7.25 Total electric power required, P = 800 kW = 800 $*{10}^3$ W

Supply voltage, V = 220 V

Electric plant generating voltage, V' = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wires lines carrying power = 0.5 Ω /km

Total resistance of the wire, R = 2 $*15*0.5\Omega$ = 15 Ω

Step-down transformer rating 4000 – 220 V, hence

Input voltage to the transformer,  $V_1$ = 4000 V

Output voltage from the transformer,  $V_2$ = 220 V

rms current in the wire lines is given as

I = $\frac{P}{V_1}$ = $\frac{800\mathrm{}*{10}^3}{4000}$ = 200 A

Line power loss = $I^{2}R$ = ${200}^2*$ 15 = 600 $*{10}^3$ W = 600 kW

Since the leakage power loss is negligible, the total power to be supplied by plat = 800 + 600 = 1400 kW

Voltage drop in the transmission line = IR = 200 $*15$ = 3000 V. Hence, the total voltage transmitted from plant = 3000 + 4000 = 7000 V

Since the power generated is 440V, the transformer rating at the plant should be 440 V – 7000 V