(i) According to the laws of photoelectric emission, the photoelectric current depends on the intensity of incident light. The constant saturation value of photoelectric current reveals that intensity of incident light is constant.
(ii) Frequency VI is the highest among v₁, v₂ and v₃ because higher the cut-off potential, higher will be the frequency of incident light.

In α-decay, the atomic number decreases by 2 units and mass number decreases by 4 units. In ß-decay, the atomic number increases by I unit but mass number does not change. In y-decay, there is no change in atomic number and mass number.
Let the mass number and atomic number of A be X and Y, respectively.
So, \(_{ Y }{ A }^{ x }\overset { \alpha }{ \longrightarrow } _{ Y-2 }{ A_{ 1 } }^{ x-4 }\overset { \beta }{ \longrightarrow } _{ Y-2+1 }{ A_{ 2 } }^{ x-4 }\)
\(\ or\ _{ Y-1 }{ A }_{ 2 }^{ x-8 }\overset { \alpha }{ \longrightarrow } _{ Y-1-2 }{ A }_{ 3 }^{ x-4-4 }\)
\(\ or\ _{ Y-3 }{ A }_{ 3 }^{ x-8 }\overset { \gamma }{ \longrightarrow } _{ Y-3 }{ A }_{ 4 }^{ x-8 }\)
According to the question, the mass number and atomic number of A₄ are 172 and 69.
∴ X - 8 = 172 ⇒ X = 172 + 8 = 180
Y - 3 = 69 ⇒ Y = 72

de-Broglie wavelength of accelerating charged particle is given by
\(\lambda =\frac { h }{ \sqrt { 2mqV } } \Rightarrow \lambda \sqrt { V } =\frac { h }{ \sqrt { 2mq } } =constant\)
(i) The slope of the lines represent \(\frac { h }{ \sqrt { 2mq } } .\)
where, h is Planck's constant, q is the charge and m is the mass of charged particle.
(ii) ₁H² and ₁H³ carry same charge (as they have same atomic number).
\(\therefore \ \lambda \sqrt { V } \propto \frac { 1 }{ \sqrt { m } } \)
The lighter mass i.e ₁H²  is represented by line of greater slope i.e A and similarly ₁H³ by line B.

\( \lambda_0=\frac{h c}{\phi_0} \quad\left[\because \phi_0=\frac{h c}{\lambda_0}\right] \\ \because \lambda_0 \propto \frac{1}{\phi_0} \)
Metal B has lower threshold wavelength.

We know that, \(\frac{1}{2} m v_{\max }^2=e V_0=h\left(v-v_0\right)\)
Here, the frequency of the incident radiation is greater than the threshold frequency. Therefore, the value of stopping potential (V₀) increases with increase in frequency (v) of the incident radiation arid KE will also increases.

The kinetic energy which is due to velocity.
From first postulate of Bohr's atom model, we have
\( \frac{m v^{2}}{r} =\frac{k Z e^{2}}{r^{2}} \)
\(\therefore \frac{1}{2} m v^{2} =\frac{1}{2} \frac{k Z e^{2}}{r} \)
(i) The kinetic energy of the electron \(=\frac{1}{2} m v^{2}\) 
\(=\frac{k Z e^{2}}{2 r}\) Potential due to the nucleus, at any point in the orbit in which electron is revolving \(=\frac{k Z e^{}}{2 r}\)
(ii) The potential energy of the electron = Potential x Charge
\(\therefore \quad \text { P.E. }=\frac{k Z e(-e)}{r}=-\frac{k Z e^{2}}{r}\)
The Lyman series is obtained when an electron jumps into first orbit (n_i = 1) from any outer orbit (n₂ = 2, 3, 4 ...)

(i) According to Bohr's principle, electrons revolve in a stationary orbit of which energy and momentum are fixed. The momentum of electrons in the fixed orbit is given by \(\frac{n h}{2 \pi}\) (where, n = number of orbits)
According to de-Broglie's hypothesis, the electron is associated with wave character. Hence, a circular orbit can be taken to be a stationary energy state only if it contains an integral number of de-Broglie wavelengths, i.e. 2\(\pi\)r = n\(\lambda\) 
(ii) According to question,
\(E_{B}-E_{C}=\frac{h c}{\lambda_{1}}\)         ............(i)
\(E_{A}-E_{B}=\frac{h c}{\lambda_{2}}\)          . ..........(ii)
\(E_{C}-E_{A}=\frac{-h c}{\lambda_{3}}\)      ..............(iii)
On adding Eqs. (i), (ii) and (iii), we get E_B - E_C + E_A - E_B + E_C - E_A
\(=h c\left(\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}-\frac{1}{\lambda_{3}}\right)\)
\(\Rightarrow \quad \frac{1}{\lambda_{3}}=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}} \Rightarrow \lambda_{3}=\frac{\lambda_{1} \lambda_{2}}{\lambda_{1}+\lambda_{2}}\)
This is the required expression.

(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

(i) Bohr's second postulate states that the electron revolves around the nucleus in certain privileged orbit which satisfy certain quantum condition that angular momentum of an electron is an integral multiple of  \(\cfrac { h }{ 2\pi } \), where h is Planck's constant. i.e  \(L=mvr=\cfrac { nh }{ 2\pi } \)
Where, m = mass of electron, v = speed of electron and r = radius of orbit of electron.
\(\therefore\) Circumference of electron in n^th orbit
= n \(\times\)de-Broglie wavelength associated with electron.  \(\left[ \because \lambda =\cfrac { h }{ mv } \right] \)            
(ii) Given, the electron in H-atom is initially in third excited state.
n = 4
And the total number of spectral lines of an atom that can exist is given by the relation
 \(=\ \cfrac { n(n-1) }{ 2 } \)  
Here,  n = 4
So, number of spectral lines = \(\cfrac { 4(4-1) }{ 2 } =\cfrac { 4\times 3 }{ 2 } =6\)
Hence, when a H-atom moves from third excited state to ground state, it emits six spectral lines. 

\(-0.85 \ eV, \ 12.75 \ eV; \ 970\mathring { A } \)
\(Here,\ { E }_{ 1 }=-13.6\ eV;\ n=4,\ { E }_{ 4 }=?\)
\( (a) \ { E }_{ 4 }=-\frac { { E }_{ 1 } }{ { 4 }^{ 2 } } =\frac { -13.6eV }{ 16 } =-0.85eV\)
(b) Energy absorbed by the atom in transition from n = 1 to n = 4 is  
\(\Delta E={ E }_{ 4 }-{ E }_{ 1 }=-0.85-(-13.6)\)
\( =12.75eV\)
\(\\ (c) \ \Delta E=\frac { hc }{ \lambda }\)
\(or \ \lambda =\frac { hc }{ \Delta E } =\frac { 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 12.75\times 1.6\times { 10 }^{ -19 } } \)
\( =970\times { 10 }^{ -10 }m\)
\(970\mathring { A } \)

\(Here, \ { \lambda }_{ 1 }=4000\overset { \circ }{ A } =4\times { 10 }^{ -7 }m;{ V }_{ 1 }=0.82V;\)
\({ \lambda }_{ 2 }=3000\overset { \circ }{ A } =3\times { 10 }^{ -7 }m;{ V }_{ 2 }=1.85V\)
\( As \ \frac { hc }{ { \lambda }_{ 1 } } ={ \phi }_{ 1 }+e{ V }_{ 1 } \ and \ \frac { hc }{ { \lambda }_{ 2 } } ={ \phi }_{ 1 }+e{ V }_{ 2 }\)
\(\therefore \ \frac { hc }{ { \lambda }_{ 2 } } -\frac { hc }{ { \lambda }_{ 1 } } =e\left( { V }_{ 2 }-{ V }_{ 1 } \right)\)
\( or \ hc\left( \frac { { \lambda }_{ 1 }-{ \lambda }_{ 2 } }{ { \lambda }_{ 1 }{ \lambda }_{ 2 } } \right) =e\left( { V }_{ 2 }-{ V }_{ 1 } \right) \)
\(or \ \ h=\frac { e\left( { V }_{ 2 }-{ V }_{ 1 } \right) { \lambda }_{ 1 }{ \lambda }_{ 2 } }{ \left( { \lambda }_{ 1 }-{ \lambda }_{ 2 } \right) c }\)
\( =\frac { \left( 1.6\times { 10 }^{ -19 } \right) \times \left( 1.85-0.82 \right) \times 4\times { 10 }^{ -7 }\times 3\times { 10 }^{ -7 } }{ \left( 4\times { 10 }^{ -7 }-3\times { 10 }^{ -7 } \right) \times 3\times { 10 }^{ 8 } }\)
\( =6.592\times { 10 }^{ -34 }Js\)

\(Here,\lambda =300nm=300\times { 10 }^{ -9 }m\)
\(=3\times { 10 }^{ -7 }m,{ V }_{ 0 }=0.54V\)
(i) Energy of the incident photon,
\(E=\frac { hc }{ \lambda } =\frac { \left( 6.63\times { 10 }^{ -34 }Js \right) \times \left( 3\times { 10 }^{ 8 }m/s \right) }{ 3\times { 10 }^{ -7 } } =6.63\times { 10 }^{ -19 }J\)
\(=\frac { 6.63\times { 10 }^{ -19 } }{ 1.6\times { 10 }^{ -19 } } eV=4.14eV\)
(ii) Max. K.E. of emitted photoelectron is
\({ K }_{ max }=e{ V }_{ 0 }=e\times 54V=0.54eV\)
(iii) \({ As\ K }_{ max }=\frac { hc }{ \lambda } -{ \phi }_{ 0 }\ or{ \ \phi }_{ 0 }=\frac { hc }{ \lambda } -{ K }_{ max }\)      
\(=4.14eV-0.54eV=3.6eV\)