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

(i) The basic concept used in the working of complexation meter is a photoelectric effect.
(ii) The reading of complexation meter depends on the current generated from the light reflected from the face. Fairer the colour more is the light reflected and vice versa.
(iii) Complexation meters judge your beauty only. The inner beauty of a person is judged by the virtues he or she posseses. In my opinion, inner beauty is much more valuable than the face beauty.

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.

\(\lambda =\frac { h }{ mv }\)
\(=\frac { 6.63\times { 10 }^{ -34 } }{ 9.1\times { 10 }^{ -31 }\times 2.2\times { 10 }^{ 8 } }\)
\(=3.31\times { 10 }^{ -12 }m\)

Given, Kinetic energy = KE = 120 eV
p=\(\sqrt { 2eVm } =\sqrt {2KE.m }\)  \([\because K E=e V]\)
\(P=\sqrt { 2\times 120\times 1.6\times 10^{ -19 }\times 9.1\times 10^{ -31 } } \)
\(=5.91\times 10^{ -24 }\ kg-m/s\)
(ii) We know that momentum, p = mv
or, \(v=\frac{p}{m}=\frac{5.91 \times 10^{-24}}{9.1 \times 10^{-31}}\)
\(=6.5 \times 10^{6} \mathrm{~m} / \mathrm{s}\)
(iii) de-Broglie wavelength associated with electron,
\(\lambda =\frac { 12.27 }{ \sqrt { V_{ } } } \mathring { A } =\frac { 12.27 }{ \sqrt { 120 } } \mathring { A=0.112\times 10^{ -9 } } \ m=0.112 \ nm\)

Let v₁ and v₂ be the velocities of the two particles of masses m₁ and m₂ respectively
According to law of conservation of linear momentum \(\ { m }_{ 1 }\overset { \rightarrow }{ { v }_{ 1 } } +{ m }_{ 2 }\overset { \rightarrow }{ { v }_{ 2 } } =M\times 0=0 \)
\( { m }_{ 1 }\overset { \rightarrow }{ { v }_{ 1 } } =-{ m }_{ 2 }\overset { \rightarrow }{ { v }_{ 2 } } \ or \ \left| { m }_{ 1 }\overset { \rightarrow }{ { v }_{ 1 } } \right| =\left| { m }_{ 2 }\overset { \rightarrow }{ { v }_{ 2 } } \right| \)
\( { \lambda }_{ 1 }=\frac { h }{ { m }_{ 1 }{ v }_{ 1 } } and{ \lambda }_{ 2 }=\frac { h }{ { m }_{ 2 }{ v }_{ 2 } } \)
\(or \ \frac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =\frac { { m }_{ 2 }{ v }_{ 2 } }{ { m }_{ 1 }{ v }_{ 1 } } =1\)

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.

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.

For a given metal, there exists certain minimum frequency of the incident radiation below which no emission of photoelectrons takes place. This frequency is called threshold frequency. It is a constant quantity for a given metal surface.

For the emission of photoelectrons from the cesium metal, the minimum energy of the incident light on the metal surface should have the photon of energy 2eV.

The work function of alkali metal is very low. Even the ordinary visible light can bring about emission of photoelectrons from these metals. Due to this reason, alkali metal surfaces are most suited as photosensitive surfaces.

(a) For the cut-off or threshold frequency, the energy h v₀ of the incident radiation must be equal to work function  Φ₀, so that
\({ V }_{ 0 }=\frac { { \phi }_{ 0 } }{ h } =\frac { 2.14eV }{ 6.63\times { 10 }^{ -34 }Js }\)
\(=\frac { 2.14\times 1.6\times { 10 }^{ -19 }J }{ 6.63\times { 10 }^{ -34 }Js } =5.16\times { 10 }^{ 14 }Hz\)
Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected.
(b) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals the potential energy eV₀ by the retarding potential V₀. Einstein’s Photoelectric equation is 
\(e{ V }_{ 0 }=\frac { hc }{ \lambda } -{ \phi }_{ 0 }\)
\(or\ \lambda =\frac { hc }{ \left( e{ V }_{ 0 }+{ \phi }_{ 0 } \right) }\)
\(or \ \lambda =\frac { \left( 6.63\times { 10 }^{ -34 }Js \right) \times \left( 3\times { 10 }^{ 8 }m/s \right) }{ \left( e\times 0.6V+2.14eV \right) } \)
\(\lambda =\frac { 19.89\times { 10 }^{ -26 }Jm }{ 2.74\times 1.6\times { 10 }^{ -19 }J } =454nm\)

M2>2M1