\({ v }_{ 0 }={ \phi }_{ 0 }/h=4.2\times 1.6\times { 10 }^{ -19 }/6.6\times { 10 }^{ -34 }\)
\(=6.72\times { 10 }^{ 15 }/6.6 \ Hz\)
\(v=c/\lambda =3\times { 10 }^{ 8 }/330\times { 10 }^{ -9 }=3\times { 10 }^{ 15 }/3.3Hz\)
\( =6\times { 10 }^{ 15 }/6.6Hz\)
As \(v<{ v }_{ 0 }\)therefore no photoelectric emission will take place.

\(p=\frac{E}{c}=\frac{3 \times 1.6 \times 10^{-13}}{3 \times 10^8}=1.6 \times 10^{-21} \mathrm{kgms}^{-1}\)

One photon can eject one photoelectron from the surface of a metal provided its energy is not less than the work function of the metal. Since each photon has energy less than the work function of copper, hence, the photoelectric emission is not possible.

The work function of a metal is the minimum energy required by an electron to just escape from the metal surface so as to overcome the restraining forces at the surface.
The relation between work function \(\left(\phi_0\right)\) and threshold frequency \(\left(v_0\right) \text { is } \phi_0=h v_0\) where h is Plank's constant. 

\(M a x . K . E .=K_{\max }=e V_0=e \times 1.5 \mathrm{~V}=1.5 \mathrm{eV}\)

1. Matter-wave is an important part of quantum physics.
2. It is said that all matter shows wave-like behavior
3. The wavelength (\(\lambda\))  of a matter-wave can be determined by is \(\lambda=\frac{h}{p}\)
where h represents Planck's constant, and p represents the traveling particle's momentum.
4. The matter wave is also called as de Broglie wave.
5. The matter-wave describes the relationship between momentum and wavelength
6. The wavelength is inversely proportional to the momentum (mass and velocity) of the particle
7. The smaller the wavelength of the matter wave, the faster the particle moves.
8. The De-Broglie wavelength increases as the particle become lighter.
\(V_{\mathrm{a}}=V_P=V\)
\(\therefore \quad \frac{\lambda_{p}}{\lambda_a}=\frac{\sqrt{2 m e V_a}}{\sqrt{2 m e V_p}}=\sqrt{\frac{V_a}{V_p}}=1\)
\(\therefore \quad \lambda_p: \lambda_a=1: 1\)

Einstein's photoelectric equation 
\(K_{\max }=\frac{1}{2} m v_{\max }^{2}=h v-\phi_{0}=h v-h v_{0}\)
The two characteristic properties of photon on which
the above equation is based are given below.
(i) A photon carries energy (hv).
(ii) One photon can eject only one electron irrespective of its energy
The following are the three observed basic features:
(i) The kinetic energy of photoelectron is independent of intensity of radiation. It depends on the frequency of radiation.
(ii) K_max must be non-negative, i.e. photoelectric emission is possible only \(\text { if } h v>\phi_{0} ; v>v_{0}\) where  is threshold frequency and \(\phi_{0}\) is work function.
(iii) The greater is the intensity of radiation the more are the numbers of photoelectrons

(i) Graph 1; Intensity
Graph 2 Frequency
(ii) Graph 1 Saturation current
Graph 2 Stopping potential
(iii) For a given photo-sensitive surface electrons need a minimum energy to be emitted, this is called work function of the surface W.
\(\therefore\) Photons energy hv should be greater/equal to the work function.
\(hv\ge w\)
\(v\ge \frac { w }{ h } \)
\(\therefore\) Minimum frequency for photo emission
\(v_{ 0 }=\frac { w }{ h } \)

(i) Since, Q has greater negative intercept, it will have greater \(\phi\) (work function) and hence higher threshold frequency.
(ii) To know work function of Q, we put
V = 0 in the following equation.
\(\begin{array}{rlrl} V =\frac{h v}{e}-\frac{\phi}{e} \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & 0 & =\frac{h v}{e}-\frac{\phi}{e} \Rightarrow \phi=h v \\ \end{array}\)
\(\begin{array}{rlrl} \therefore & \phi & =6.6 \times 10^{-34} \times 6 \times 10^{14} \mathrm{~J} \end{array}\)
\(=\frac{6.6 \times 6 \times 10^{-20}}{1.6 \times 10^{-19}} \mathrm{eV}=2.5 \mathrm{eV}\)
(iii) From the equation, \(v \lambda=c\)
\(\begin{aligned} \Rightarrow \quad \lambda & =\frac{c}{v}=\frac{3 \times 10^8}{8 \times 10^{14}}=\frac{30}{8} \times 10^{-7} \mathrm{~m} \end{aligned}\)
\(\begin{aligned} =\frac{30}{8} \times 10^3 \times 10^{-10} \mathrm{~m} \end{aligned}\)
\(\begin{aligned} =\frac{30}{8} \times 10^3 \end{aligned}\) \(\overset{\circ}{A}\)= 3750 \(\overset{\circ}{A}\)
Energy \(=\frac{12375}{\lambda(\overset{\circ}{A})} \mathrm{eV}=\frac{12375}{3750} \mathrm{eV}=33 \mathrm{eV}\)
\(\therefore\) Maximum KE of emitted electron = 33 - 2.5 eV
= 0.8 eV

de-Broglie matter wave equation,
\(\lambda =\frac { h }{ p } =\frac { h }{ \sqrt { 2mK } } \ \left[ \because K=\frac { { P }^{ 2 } }{ 2m } \right] \)
where K is kinetic energy and m is a mass of the particle.
\(K=\frac { { h }^{ 2 } }{ 2m{ \lambda }^{ 2 } } \)   [ for same wavelength \( \lambda] \)
\(K\propto \frac { 1 }{ m }\)
\(\Rightarrow \ { K }_{ e }:{ K }_{ \alpha }:{ K }_{ p }=\frac { 1 }{ { m }_{ e } } :\frac { 1 }{ { m }_{ \alpha } } :\frac { 1 }{ { m }_{ p } } \)
where \({ m }_{ e },{ m }_{ p } \ and \ { m }_{ \alpha }\) are masses of electron, proton and \(\alpha \) -particle, respectively.
Also, \({ K }_{ e },k_{ p } \ and \ K_{ \alpha }\) are their respective kinetic energies.
\(\because \ m_{ \alpha }>m_{ p }>m_{ e }\)
\(\\ \Rightarrow m_{ \alpha }m_{ p }>m_{ e }m_{ \alpha }>m_{ e }m_{ p }\)
\(\\ { K }_{ e }>k_{ p }>K_{ \alpha }\)
(i) \(\alpha \) particle possess minimum kinetic energy
(ii) The electron has maximum kinetic energy. The magnifying power of an electron microscope is inversely related to the wavelength of radiation used. The Smaller wavelength of the electron beam in comparison to visible light increases the magnifying power of the microscope.

(i) Given, mass of bullet, m = 0.040 kg
Speed of bullet, v = 1 km/s = 1 \(\times\) 10³ m/s
de-Broglie wavelength,
\(\lambda=\frac{b}{m v}=\frac{6.63 \times 10^{-34}}{0.040 \times 1 \times 10^3}\)
= 1.66 \(\times\) 10^-35 m
(ii) Mass of the ball, m = 0.060 kg and speed of the ball, v=1 m/s
\(\begin{aligned} \lambda & =\frac{b}{m v}=\frac{6.63 \times 10^{-34}}{0.060 \times 1} \end{aligned}\)
\(\begin{aligned} =1.1 \times 10^{-32} \mathrm{~m} \end{aligned}\)
(iii) Mass of a dust particle, m = 1 \(\times\) 10^-9 kg and speed of the dust particle, v = 2.2 m/s
\(\lambda=\frac{b}{m v}=\frac{6.63 \times 10^{-34}}{1 \times 10^{-9} \times 2.2}\)
= 3.0 \(\times\) 10^-25 m

\(1.227\mathring { A } \quad \)

\(3\times { 10 }^{ 4 }photons\)

\(Here \ { \phi }_{ 0 }=1.8eV,\lambda =4500 \ A=4.5\times { 10 }^{ -7 }m\)
(i) Max K.E of emitted photoelectron is
\({ K }_{ max }=\frac { hc }{ \lambda } -{ \phi }_{ 0 }\)
\( =\frac { (6.6\times { 10 }^{ -34 })(3\times { 10 }^{ 8 }) }{ 4.5\times { 10 }^{ -7 } } -1.8\times 1.6\times { 10 }^{ -19 }\)
\( =4.4\times { 10 }^{ -19 }-2.88\times { 10 }^{ -19 }=1.52\times { 10 }^{ -19 }J\)
(ii)Max. velocity of emitted photoelectron
\({ v }_{ max }=\sqrt { \frac { { 2K }_{ max } }{ m } } =\sqrt { \frac { 2\times 1.52\times { 10 }^{ -19 } }{ 9.1\times { 10 }^{ -31 } } }\)
\( =5.78\times { 10 }^{ 5 }{ ms }^{ -1 }\)
(iii) The kinetic energy of the emitted photoelectron is an incident of the intensity of the incident light. Hence, if the intensity of incident light is doubled the max. K.E of the emitted photoelectron electrons remains unchanged.

(i) (b): \(K_{\max }=h v-\phi_{0}=\frac{h c}{\lambda}-\phi_{0}\)
\(=\frac{\left(6 \cdot 6 \times 10^{-34}\right) \times\left(3 \times 10^{8}\right)}{\left(300 \times 10^{-9}\right) \times\left(1 \cdot 6 \times 10^{-19}\right)}-3 \cdot 3\)
\(=4 \cdot 125 \times 3 \cdot 3=0 \cdot 825 \mathrm{eV}\)
(ii) (c) : \(K_{\max }=h v-\phi_{0}, \text { When } v=v_{0}, K_{\max }=0\)
\(\therefore \quad 0=h v_{0}-\phi_{0} \text { or } \phi_{0}=h v_{0}\)
If v < v₀ then K_max is negative, i.e., no photoelectric emission takes place. Thus, graph (c) is possible.
(iii) (a): Photoelectric current (i) is proportional to the intensity of the emission light. Thus, graph (a) is possible.
(iv) (c): From Einstein's photoelectric equation,
\(K_{\max }=e V_{0}=\frac{h c}{\lambda}-\phi\)
or \(V_{0}=\frac{h c}{e} \cdot \frac{1}{\lambda}-\frac{\phi}{e}\)
Graph of V₀ versus \(\frac{1}{\lambda}\) is a straight line Slope of straight line,\(\tan \theta=\frac{h c}{e}\)
\(\text { At } V_{0}=0, \text { we have }\)
\(\phi_{1}: \phi_{2}: \phi_{3}=\frac{h c}{\lambda_{01}}: \frac{h c}{\lambda_{02}}: \frac{h c}{\lambda_{03}}\)
\(0.001 h c: 0.002 h c: 0.004 h c\)
\(\therefore 1: 2: 4\)
(v) (d): de-Broglie wavelength
\(\lambda=\frac{h}{p} i . e ., \lambda \propto \frac{1}{p}\)
So the graph between \(\lambda\) and P is of the type shown is option (d).

(b): Stopping potential is a measure of maximum kinetic energy of emitted photoelectron (eV₀ = K_max) and K_max depends upon the frequency of incident light but is independent of intensity.