(a) Perseverance, not giving up and patience.
(b) \(\lambda =\frac { 12.27 }{ \sqrt { V } } \)\(\overset { o }{ A } \)

(i) Davisson and Germer observed diffraction patterns of slow moving electrons. And G.P. Thomson observed a diffraction pattern of fast moving electrons. As diffraction is essentially a wave phenomenon, therefore, it was concluded that wave must be associated with moving electrons.
(ii) (a) At rest, v = 0, \(=\frac { h }{ mv } =\frac { h }{ m\times 0 } =\infty \)
(b) Particle moving with infinite speed, v = \(\infty \) 
\(\lambda =\frac { h }{ mv } =\frac { h }{ m\times \infty } \)
(iii) The dual nature of moving material particles reveals in a way the nature of Almighty God. He is in a visible form (i.e., Sakar) like a visible particle and also without any form (i.e., Nirakar) like a wave.It depends on us how we realize him. ​

Since the electron has been localised within the nucleus (of size about \({ 10 }^{ -14 }\) m)of an atom, so \({ \lambda =10 }^{ -14 }\)
Momentum of electron,
  \(p=\frac { h }{ \lambda } =\frac { 6.63\times { 10 }^{ -34 } }{ { 10 }^{ -14 } } =6.63\times { 10 }^{ -20 }kg{ ms }^{ -1 }\)
The relativistic relation for the energy of an electron is
\(E=\sqrt { { p }^{ 2 }{ c }^{ 2 }+{ { m }_{ 0 } }^{ 2 }{ c }^{ 4 } } \)
Neglecting the rest-mass energy term(i.e., second term), we have
 \(E=pc=(6.63\times { 10 }^{ -20 })\times (3\times { 10 }^{ 8 })J\)
\( =\frac { 6.63\times 3\times { 10 }^{ -12 } }{ 1.6\times { 10 }^{ -13 } } MeV=124.3MeV\)
This energy E(=124.3MeV) is very large as compared to the binding energy which is provided by Coulomb's force within the nucleus. Since the energy of electron inside the nucleus, hence the electron does not reside the nucleus.

(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  \(\left(e V=\frac{1}{2} m v^{2}\right) \text { and }\left(e B v=\frac{m v^{2}}{r}\right)\)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 velocity of an electron as \(\left(v=\sqrt{2 v\left(\frac{e}{m}\right)}\right. \text { and }\left(v=B r\left(\frac{e}{m}\right)\right)\) respectively.
(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. At 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 (λ) is significant, but the frequency (ν) associated with an electron has no direct physical significance. Therefore, the product νλ (phase speed) has no physical significance.
Group speed is given as:
\(V_{G}=\frac{d v}{d k} \)
\(=\frac{\mathrm{d} \mathrm{v}}{d\left(\frac{1}{\lambda}\right)}=\frac{\mathrm{d} \mathrm{E}}{\mathrm{dp}}=\frac{d\left(\frac{p^{2}}{2} m\right)}{\mathrm{dp}}=\frac{p}{m}\)
This quantity has a physical meaning.

Potential of an anode, V = 100 V
Magnetic field experienced by the electrons, B = 2.83 x 10⁻⁴ T
Radius of the circular orbit r = 12.0 cm = 12.0 x 10⁻² m
Mass of each electron = m
Charge on each electron = e
Velocity of each electron = v
The energy of each electron is equal to its kinetic energy, i.e.,
\(\frac{1}{2} m v^{2}=e V\)
\(v^{2}=\frac{2 e V}{m}\)
It is the magnetic field, due to its bending nature, that provides the centripetal force
\(\left(F=m \frac{v^{2}}{r}\right)\) for the beam. Hence, we can write:
Centripetal force = Magnetic force
\(m \frac{v^{2}}{r}=e v B\)
\(e B=m \frac{v}{r}\)
\(v=\frac{e B r}{m}\)
Putting the value of v in equation (1), we get:
\(\frac{2 \mathrm{eV}}{m}=\frac{e^{2} B^{2} r^{2}}{m^{2}}\)
\(\frac{e}{m}=\frac{2 \mathrm{~V}}{B^{2} r^{2}}\)
\(=\frac{2 \times 100}{\left(2.83 \times 10^{-4}\right)^{2} \times\left(12 \times 10^{-2}\right)^{2}}=1.73 \times 10^{11} \mathrm{Ckg}^{-1}\)
Therefore, the specific charge ratio (e/m) is = 1.73 x 10¹¹Ckg^-1.

During photelectric emission, the momentum of incident photon is transferred to the metal. At microscopic level, atoms of a metal absorb the photon and its momentum is transferred mainly to the nucleus and electrons.
The excited electron is emitted. Therefore, the conservation of momentum is to be considerd as the momentum of incident photon transferred to the nucleus and electrons.

The following features observed in photoelectric effect helped to establish the photon picture of the electromagnetic radiation.
(i) The maximum kinetic energy of the emitted photoelectron is independent of the intensity of the incident light but depends upon the frequency of the incident light.
(ii) For every metal, there is a certain minimum frequency of the incident light below which, no photoelectric emission takes place.
(iii) The photoelectric emission is an instantaneous process.

If we increase the wavelength of incident light then the energy of incident photon \(E=\frac { hc }{ \lambda } \)will decrease. Due to it, the energy of photoelectron emitted decreases, but the number of electrons emitted will not change. When the wavelength of the incident light becomes greater than the threshold wavelength then no photoelectron will be emitted.

The maximum kinetic energy of the emitted photoelectron from a metal surface is given by 
\({ K }_{ max }=\frac { 1 }{ 2 } { mv }_{ max }^{ 2 }=\frac { hc }{ \lambda } -\ \frac { hc }{ { \lambda }_{ 0 } } =hc\left( \frac { { \lambda }_{ 0 }-\lambda }{ \lambda { \lambda }_{ 0 } } \right) \)
(i)When is \(\lambda >{ \lambda }_{ 0 }\)\({ K }_{ max }\) negative is imaginary. Thus photoelectric emission will not occur.
(ii)When \(\lambda <{ \lambda }_{ 0 }\)\({ K }_{ max }\) is positive, \({ v }_{ max }\) is positive.Thus photoelectric emission will take place and K.E. of photoelectron increases as \(\lambda \)decreases.

\( \theta=90^{\circ}-\frac{\phi}{2}=90^{\circ}-\frac{52^{\circ}}{2} \\ =64^{\circ} \)

As we know, de-Broglie wavelength \(\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m K}}\)
\(K_1=\frac{h^2}{2 m \lambda_1^2}\)
If according of the questions
\( K_2 =\frac{K_1}{4} \\ K_2 =\frac{h^2}{2 \lambda_2^2} \\ \frac{K_1}{4} =\frac{h^2}{2 m \lambda_2^2} \)
From Eqs. (i) and (ii), we get
\( \frac{\frac{K_1}{4}}{K_1}=\frac{h^2}{2 m \lambda_2^2} \times \frac{2 m \lambda_1^2}{h^2} \\ \frac{1}{4}=\frac{\lambda_1^2}{\lambda_2^2} \\ \Rightarrow \quad \frac{\lambda_1}{\lambda_2}=\frac{1}{2} \)
Hence, wavelength of the particle double the wavelength when kineettiic energy is 1/4^th.

\(I \alpha \frac{I}{r^2}\)
\(\frac{I_1}{I_2}=\left(\frac{r_2}{r_1}\right)^2\)

The value of stopping potential in photoelectric emission from a metal surface decreases with the decrease in frequency of incident radiation, provided the frequency of radiation is greater than a threshold frequency.
When the frequency of incident radiations becomes equal to a threshold frequency, stopping potential becomes zero.