Physics

In the equation $v=\sqrt{\frac{\gamma P}{\rho }}$ ……(i)

$\rho$ Density = $\frac{Mass}{Volume}$ = $\frac{M}{V}$ where M = molecular weight of the gas, V = Volume of the gas, so we can write

$v=\sqrt{\frac{\gamma PV}{M}}$ …….(ii)

For ideal gas equation, PV = nRT, n = 1 so PV = RT

For constant T, PV = constant

In equation (ii), since PV = constant, $\gamma$ and M constant, v is also constant. Hence, at a constant temperature, the speed of sound in a gaseous medium is independent of the change in the pressure of the gas.

From equation (i) $v=\sqrt{\frac{\gamma P}{\rho }}$

For 1 mole of an ideal gas, the gas equation can be written as PV = RT or P = $\frac{RT}{V}$

Substituting in equation (i), we get $v=\sqrt{\frac{\gamma RT}{\rho V}}$ = $\sqrt{\frac{\gamma RT}{M}}$

Since $\gamma$ , R and M are constant, we get v $\propto \surd T$

Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium. i.e. the speed of sound increase with an increase in the temperature of the gaseous medium and vice versa.

Let $v_m$ and $v_d$ be the speed of sound in moist air and dry air respectively and ${\rho }_m$ and ${\rho }_d$ be the corresponding densities

From equation (i) $v=\sqrt{\frac{\gamma P}{\rho }}$ , we get $v_m\mathrm{}=\sqrt{\frac{\gamma P}{{\rho }_m}}$ and $v_d\mathrm{}=\sqrt{\frac{\gamma P}{{\rho }_d}}$

$\frac{v_m}{v_d}$ = $\sqrt{\frac{{\rho }_d}{{\rho }_m}}$

However, the presence of water vapour reduces the density of air, i.e. ${\rho }_d<$ ${\rho }_m$ , so $v_m<$ $v_d$

11.37 (a) Quarks inside protons and neutrons carry fractional charges. This is because nuclear force increases extremely if they are pulled apart. Therefore, fractional charges may exist in nature; observable charges are still the integral multiple of an electrical charge.

(b) The basic relations for electric field and magnetic field are

$\left\{eV=\frac{1}{2}m{v}^2\right\}$ and $\left\{eBV=\frac{m{v}^2}{r}\right\}$  respectively

These relations include e (electric charge), v (velocity), m (mass), V (potential), r (radius) and B (magnetic field. These relations give the value of the velocity of an electron as

$\left(v=\sqrt{2V\left(\frac{e}{m}\right)}\right)and\left(v=Br\left(\frac{e}{m}\right)\right)$ respectively

It can be observed from these relations that the dynamics of an electron is determined not by e and m separately, but by the ratio of  . $\frac{e}{m}$

(c) At atmospheric pressure, the ions of gases have no chance of reaching their respective electrons because of collision and recombination with other gas molecules. Hence, gases are insulators at atmospheric pressure. A low pressures, ions have a chance of reaching their respective electrodes and constitute a current. Hence, they conduct electricity at these pressures.

(d) The work function of a metal is the minimum energy required for a conduction electron to get out of the metal surface. All the electrons in an atom do not have the same energy level. When a ray having some photon energy is incident on a metal surface, the electrons come out from different levels with different energies. Hence, these emitted electrons show different energy distributions

(e) The absolute value of energy of a particle is arbitrary within the additive constant. Hence, wavelength( $\lambda )$ ( is significant, but the frequency( $\nu )$ (  associated with an electron has no direct physical significance.

Therefore, the product $\nu \lambda$  (phase speed) has no physical significance.

Group speed is given as:

$v_G$ = $\frac{dv}{dk}$ = $\frac{dv}{d\left(\frac{1}{\lambda }\right)}$ = dE/dp = (d(p^2/2m))/dp = p/m

This quantity has a physical meaning.

11.35 Room temperature, T = 27 $?$  = 300 K

Atmospheric pressure, P = 1 atm = 1.01 $*{10}^5$  Pa

Atomic weight of helium atom = 4

Avogadro's number, $N_A$  = 6.023 $*{10}^{23}$

Boltzmann's constant, k = 1.38 $*{10}^{-23}$  J ${mol}^{-1}{K}^{-1}$

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

Average energy of a gas at temperature T is given as:

E = $\frac{3}{2}$ kT = $\frac{3}{2}*$ 1.38 $*{10}^{-23}*300$  = 6.21 $*{10}^{-21}$ J

De Broglie wavelength is given as

$\lambda =$ $\frac{h}{\sqrt{2mE}}$  , where m = mass of He atom = $\frac{Atomicweight}{Avogadr{o}^{\text{'}}snumber}$  = $\frac{4}{6.023*{10}^{23}}$

m = 6.641 $*{10}^{-24}$  gm = 6.641 $*{10}^{-27}$ kg

$\lambda =$ $\frac{6.626*{10}^{-34}}{\sqrt{2*6.641*{10}^{-27}*6.21*{10}^{-21}}}$ = 7.29 $*{10}^{-11}$ m

We have ideal gas formula:

PV = RT

PV = kNT

$\frac{V}{N}=\frac{kT}{P}$          

Where, V = volume of the gas

N = number of moles of the gas

Mean separation between two atoms of the gas is given by the relation:

r = $({\frac{V}{N})}^{1/3}$   = ${\left(\frac{kT}{P}\right)}^{1/3}$

=( ${\frac{1.38*{10}^{-23}*300}{1.01*{10}^5})}^{1/3}$

= 3.45 $*{10}^{-9}$ m

Hence, the mean separation between the atoms is much greater than the De Broglie wavelength.

11.33 The accelerating voltage of the electrons, V = 50 kV = 50 $*{10}^3$  V

Mass of the electron, $m_e$ = 9.11 $*{10}^{-31}$  kg

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

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

Wavelength of yellow light, $\lambda$ = 5.9 $*{10}^{-7}$ m

The kinetic energy of the electron is given as, $E_k$  = e $*V$  = 1.6 $*{10}^{-19}*$ 50 $*{10}^3$  J = 8 $*{10}^{-15}$ J

De Broglie wavelength is given by the relation,

$\lambda =\frac{h}{\sqrt{2*m_e*E_k\mathrm{}}}$ = $\frac{6.626*{10}^{-34}}{\sqrt{2*9.11*{10}^{-31}*8*{10}^{-15}}}$  = 5.5 $*{10}^{-12}$  m

This wavelength is nearly ${10}^5$ times less than the wavelength of a yellow light.

The resolving power of a microscope is inversely proportional to the wavelength of light used. Thus, the resolving power of an electron microscope is nearly ${10}^5$ times that of an optical microscope.