Dual Nature of Radiation and Matter

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

This wavelength is comparable to the inter-atomic spacing of a crystal ( ${10}^{-10}$ m). Hence, the high energy neutron beam should first be thermalised, before using for diffraction.

11.30 Intensity of the incident light, I = ${10}^{-5}$  W $m^{-2}$

Surface area of the sodium photocell, A = 2 ${cm}^2$  = 2 $*{10}^{-4}{m}^2$

The incident power of the light, P = I $*A$  = ${10}^{-5}$ $*{10}^{-4}{m}^2$ W = 2 $*{10}^{-9}$  W

Work function of the metal, ${\varnothing }_0$ = 2 eV = 2 $*1.6*{10}^{-19}$  J = 3.2 $*{10}^{-19}$  J

Number of layers of sodium that absorbs the incident energy, n = 5

The effective atomic area of a sodium atom, $A_e$  = ${10}^{-20}$ $m^2$

Hence, the number of conduction electrons in n layers is given by:

n' = n $*\frac{A}{A_e\mathrm{}}$  = 5 $*\frac{2*{10}^{-4}}{{10}^{-20}}$  = 1 $*{10}^{17}$

Since the incident power is uniformly absorbed by all the electrons continuously, the amount of energy absorbed per second per electron is

$E=\frac{P}{n\text{'}}$  = $\frac{2*{10}^{-9}}{1*{10}^{17}}$  = 2 $*{10}^{-26}$  J/s

Time required for photoelectric emission,

t = $\frac{{\varnothing }_0}{E}$  = $\frac{3.2*{10}^{-19}}{2*{10}^{-26}}$ s = 16 $*{10}^6$  seconds = 0.507 years

The time required for the photoelectric emission is nearly half a year, which is not practical. Hence, the wave picture is in disagreement with the given experiment.

11.29 Wavelength of the radiation, $\lambda$ = 3300 Å = 3300 $*{10}^{-10}$ m

Speed of light, c = 3 $*{10}^8$ m/s

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

The energy of the incident radiation is given as:

$E$ = $\frac{hc}{\lambda }$

= $\frac{6.626*{10}^{-34}*3*{10}^8\mathrm{}}{3300*{10}^{-10}}$ = 6.024 $*{10}^{-19}$ J = $\frac{6.024*{10}^{-19}}{1.6*{10}^{-19}}$ eV = 3.765 eV