Physics

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

It can be observed that the energy of incident radiation is greater than the work function of Na and K only. Mo and Ni having more work function will not show the photoelectric emission.

If the source of light is brought near the photocells and placed 50 cm away from them, then the intensity of radiation will increase, but it will not affect the energy of the radiation. Hence, the result will be the same as before. However, the photoelectrons emitted from Na and K will increase in proportion to intensity.