Ncert Solutions Physics Class 12th

9.6 The angle of minimum deviation, ${\delta }_m$ = 40 $°$

Angle of prism, A = 60 $°$

Let the refractive index of water, $=1.33$ , and the refractive index of prism material = $\mu \text{'}$

The angle of deviation is related to refractive index $\mu \text{'}$ is given as

${\mu }^{\text{'}}=\frac{\sin ?\frac{(A+{\delta }_m)}{2}}{\sin ?\frac{A}{2}}$ = $\frac{\sin ?\frac{(60°+40°)}{2}}{\sin ?\frac{60°}{2}}$ = $\frac{\sin ?50°}{\sin ?30°}$ = 1.532

So the refractive index of prism material is 1.532

Since the prism is placed in water, let ${\delta }_m^{\text{'}}$ be the new angle of minimum deviation.

The refractive index of glass with respect to water is given by the relation:

${\mu }_g^w$ = $\frac{{\mu }^{\text{'}}}{\mu }$ = $\frac{\sin ?\frac{(A+{\delta }_m^{\text{'}}\mathrm{})}{2}}{\sin ?\frac{A}{2}}$

$\frac{1.532}{1.33}$ = $\frac{\sin ?\frac{(60°+{\delta }_m^{\text{'}}\mathrm{})}{2}}{\sin ?\frac{60°}{2}}$

1.152 $*\sin ?30°$ = $\sin ?\frac{(60°+{\delta }_m^{\text{'}}\mathrm{})}{2}$

$\frac{(60°+{\delta }_m^{\text{'}}\mathrm{})}{2}={\sin }^{-1}?0.576$ = 35.2 $°$

${\delta }_m^{\text{'}}=2*$ 35.2 $°$ - 60 $°$ = 10.33 $°$

Hence the new minimum angle of deviation is 10.33$°$

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.

11.32 Kinetic energy of neutron, $E_{kn}$ == 150 eV = 150 $*1.6*{10}^{-19}$ J = 2.4 $*{10}^{-27}$ J

Mass of a neutron, $m_n$ = 1.675 $*{10}^{-27}$ kg

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

Kinetic energy of a neutron is given by the relation:

 = $E_{kn}=\frac{1}{2}{m}_n{v}_n^2$……….(1)

Where, $v_n$ is the velocity of neutron

The de Broglie wavelength is given by

$\lambda =\frac{h}{p}$ , where p = momentum = $m_n{v}_n$

$\lambda =\frac{h}{m_n{v}_n}$

$v_n$ = $\frac{h}{m_n\lambda }$

Substituting the value of  $v_n$ in equation (1), we get

$E_{kn}=\frac{1}{2}{m}_n{\left(\frac{h}{m_n\lambda }\right)}^2$ = $\frac{h^2}{2{m_n\lambda }^2}$

  ${\lambda }^2$ = $\frac{h^2}{2m_n{E}_{kn}}$

$\lambda =$ $\sqrt{\frac{h^2}{2m_n{E}_{kn}}}$ = $\frac{h}{\sqrt{2m_n{E}_{kn}}}$ = $\frac{6.626*{10}^{-34}}{\sqrt{2*1.675*{10}^{-27}*2.4*{10}^{-17}}}$  = 2.337 $*{10}^{-12}$ m

 It is given in the previous problem that the inter-atomic spacing of a crystal is about 1 Å, i.e. ${10}^{-10}$ m. Hence, the inter-atomic spacing is about ${10}^2$  times greater. Hence, a neutron beam of energy 150 eV is not suitable for diffraction experiments.

Room temperature, T = 27 $?$ = 27 + 273 = 300 K

The average kinetic energy is given by the equation, $E_{kn}$ = $\frac{3}{2}$ kT

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

The wavelength of neutron is given by

$\lambda =$ $\sqrt{\frac{h^2}{2m_n{E}_{kn}}}$ = $\frac{h}{\sqrt{2m_n{E}_{kn}}}$ = $\frac{h}{\sqrt{2m_n*\frac{3}{2}\mathrm{k}\mathrm{T}}}$ = $\frac{h}{\sqrt{3m_{n}kT}}$

= $\frac{6.626*{10}^{-34}}{{(3*1.675*{10}^{-27}*1.38*{10}^{-23}*300)}^{\frac{1}{2}}}$  = 1.45 $*{10}^{-10}$ m