Dual Nature of Radiation and Matter

11.25 The power of the medium wave transmitter, P = 10 kW = 10 $*{10}^3$ W = ${10}^4$ J/s

Hence energy emitted by the transmitter per second, E = ${10}^4$ J

Wavelength of the radio wave, $\lambda$ = 500 m

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

Speed of light, c = 3 $*{10}^8$ m/s

Energy of the wave is given as :

$E_w$ = $\frac{hc}{\lambda }$ = $\frac{6.626*{10}^{-34}*3*{10}^8}{500}$ = 3.98 $*{10}^{-28}$ J

Let n be number of photons emitted by the transmitter. Hence, total energy transmitted is given by:

n $E_w$ = E

n = $\frac{E}{E_w}$ = $\frac{{10}^4}{3.98*{10}^{-28}}$ = 2.52 $*{10}^{31}$

Intensity of light perceived by the human eye, I = ${10}^{-10}$ W $m^{-2}$

Area of the pupil, A = 0.4 ${cm}^2$ = 0.4 $*{10}^{-4}{m}^2$

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

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

Energy emitted by photon is given as:

$E=h\nu$ = 6.626 $*{10}^{-34}*6*{10}^{14}$ = 3.9756 $*{10}^{-19}$ J

If n is total number of photons falling per second, per unit area of the pupil,

Total energy per unit area of n photons is = n $*$ 3.9756 $*{10}^{-19}$ J = Intensity of light

n $*$ 3.9756 $*{10}^{-19}$ = ${10}^{-10}$

n = 2.52 $*{10}^8$

Total number of photons entering the pupil per second is given as:

$n_A$ = n $*A$ = 2.52 $*{10}^8*$ 0.4 $*{10}^{-4}$ = 10.06 $*{10}^3$ /s

The number is not as large as the one found in problem (a), but it is large enough for the human eye to never see the individual photons.

11.21 Speed of the electron, v = 5.20 $*{10}^{6}m/s$

Magnetic field experienced by the electron, B = 1.30 $*{10}^{-4}$ T

Specific charge of electron, e/m = 1.76 $*{10}^{11}$ C/kg

Charge of an electron e = 1.60 $*{10}^{-19}$ C

Mass of electron, m = 9.1 $*{10}^{-31}$ kg

The force exerted on the electron is given as

F = $e\left|\stackrel{?}{v}+\stackrel{?}{B}\right|$

= $evB\mathit{sin}?\theta$ , where $\theta$ = angle between the magnetic field and the beam velocity

The magnetic field is normal to the direction of beam, hence $\theta =90°$

Therefore, $F=evB$ ………………(1)

The beam traces a circular path of radius r. The magnetic field due to its bending nature provides a centrifugal force (F = $\frac{m{v}^2}{r}$ ) for the beam.

Therefore, $evB$ = $\frac{m{v}^2}{r}$

r = $\frac{m{v}^2}{evB}$ = $\frac{mv}{eB}$

= $\frac{9.1*{10}^{-31}*5.20*{10}^6}{1.60*{10}^{-19}*1.30*{10}^{-4}}$ = 0.2275 m = 22.75 cm

Energy of the electron beam, E = 20 MeV = 20 $*{10}^6$ eV = 20 $*{10}^6*1.6*{10}^{-19}$ J

E = 3.2 $*{10}^{-12}$ J

The energy of the electron beam is given as:

E = $\frac{1}{2}$ m $v^2$

v = $\sqrt{\frac{2E}{m}}$ = $\sqrt{\frac{2*3.2*{10}^{-12}}{9.1*{10}^{-31}}}$ = 2.652 $*{10}^9$ m/s

This is not possible nothing can move faster than the light. In the above formula

$E_{kinetic}$ = $\frac{1}{2}m{v}^2$ can only be used in the non-relativistic limit, i.e. v << c

For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the mass is given as :

$m=m_0\sqrt{(1-\frac{v^2}{c^2}})$ where $m_0$ = Mass of the particle at rest.

Hence, the radius of the circular path is given as

r = $\frac{mv}{eB}$ = $\frac{v}{eB}*m_0\sqrt{(1-\frac{v^2}{c^2}})$

11.20 Potential difference across evacuated tube, V = 500 V

Specific charge of electron, e/m = 1.76 $*{10}^{11}$ C/kg

The speed of each emitted electron is given by the relation of kinetic energy as

$E_{kinetic}$ = $\frac{1}{2}m{v}^2$ = $eV$

$v=\sqrt{\frac{2eV}{m}}$ = $\sqrt{2V\frac{e}{m}}$ = $\sqrt{2*500*1.76*{10}^{11}}$ = 13.27 $*{10}^6$ m/s

Therefore, the speed of each emitted electron is 13.27 $*{10}^6$ m/s

Collector potential, V = 10 MV = 10 $*{10}^6$ V

The speed is given by $v=\sqrt{\frac{2eV}{m}}$ $=\sqrt{2V\frac{e}{m}}$ = $\sqrt{2*10\mathrm{}*{10}^6\mathrm{}*1.76*{10}^{11}}$ = 1.876 $*{10}^9$ m/s

This is not possible nothing can move faster than the light. In the above formula

$E_{kinetic}$ = $\frac{1}{2}m{v}^2$ can only be used in the non-relativistic limit, i.e. v << c

For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the total energy is given as :

$E=m{c}^2$ , where m = relativistic mass

$E=m_0\sqrt{(1-\frac{v^2}{c^2}})$ where $m_0$ = Mass of the particle at rest.

Hence, kinetic energy is given as

$E_{kinetic}$ = m $c^2$ - $m_0{c}^2$

11.19 Temperature of the Nitrogen molecule, T = 300 K

Atomic mass of nitrogen = 14.0076 u

Hence, mass of Nitrogen molecule, m = 2 $*$ 14.0076 u = 28.0152 u

We know, 1 u = 1.66 $*{10}^{-27}$ kg

So, m = 28.0152 $*$ 1.66 $*{10}^{-27}$ kg = 4.65 $*{10}^{-26}$ kg

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

Boltzmann constant, k = 1.38 $*{10}^{-23}$ kg $m^2{s}^{-2}{K}^{-1}$

We have the expression that relates to mean kinetic energy ( $\frac{3}{2}kT)$ of the nitrogen molecule with root mean square speed ( $v_{rms}$ ) as:

$\frac{1}{2}m{v_{rms}}^2$ = $\frac{3}{2}kT$

$v_{rms}=\sqrt{\frac{3kT}{m}}$ = $\sqrt{\frac{3*1.38*{10}^{-23}*300}{4.65*{10}^{-26}}}$ = 516.814 m/s

De Broglie wavelength of the nitrogen molecule

$\lambda =\frac{h}{m{v}_{rms}}$ = $\frac{6.626\mathrm{}*{10}^{-34}}{4.65*{10}^{-26}*516.814}$ = 2.76 $*{10}^{-11}$ m = 0.0276 nm

Therefore, De Broglie wavelength is 0.0276 nm.

11.17 De Broglie wavelength of the neutron, $\lambda$ = 1.40 $*{10}^{-10}$ m

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

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

Kinetic energy, $E_k$ = $\frac{1}{2}m{v}^2$ ………….(1)

De Broglie wavelength and velocity (v) are related as

$\lambda =\frac{h}{mv}$ …………….(2)

Combining equation (1) and (2), we get

$E_k$ = $\frac{1}{2}m({\frac{h}{m\lambda })}^2$ = $\frac{h^2}{2m{\lambda }^2}$ = $\frac{{(6.626\mathrm{}*{10}^{-34})}^2}{2*1.66\mathrm{}*{10}^{-27}*({1.40\mathrm{}*{10}^{-10})}^2}$ = 6.75 $*{10}^{-21}$ J = $\frac{6.75*{10}^{-21}}{1.6*{10}^{-19}}$ eV = 42.17 $*{10}^{-3}eV$

Temperature of the neutron, T = 300 K

Average kinetic energy of the neutron, $E_{k-avg}$ = $\frac{3}{2}kT$ ,

where k = Boltzmann constant = 1.38 $*{10}^{-23}$ kg $m^2{s}^{-2}{K}^{-1}$

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