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.
\(V_{\mathrm{a}}=V_P=V\)
\(\therefore \quad \frac{\lambda_{p}}{\lambda_a}=\frac{\sqrt{2 m e V_a}}{\sqrt{2 m e V_p}}=\sqrt{\frac{V_a}{V_p}}=1\)
\(\therefore \quad \lambda_p: \lambda_a=1: 1\)

The energy of incident photon is calculated as 
\( E =\frac{h c}{\lambda} \)
\(=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{2.5 \times 10^{-7} \times 1.6 \times 10^{-19}}=4.9 \mathrm{eV} \)
\(\text { (i) } \because \quad E_{k(\max )} =\frac{h c}{\lambda}-\phi_{0}=4.9-3.5=1.4 \mathrm{eV} \)
(ii) The slowest electrons emitted from the surface will have zero kinetic energy.
As the work function of metal surface (5.5 eV) is more than the energy of incident photon, it will not emit any photoelectron.

It is due to the fact that photoelectric emission takes place only above threshold frequency. As frequency of red light is less than this threshold frequency, it cannot cause photoelectric emission whatsoever be the intensity.
We know that, \(\frac{h c}{\lambda}=\frac{h c}{\lambda_{0}}+\frac{1}{2} m v^{2}\)
On neglecting work function, we get
\( \frac{h c}{\lambda}=\frac{1}{2} m v^{2} \)
\(\therefore m v=\sqrt{\frac{2 m h c}{\lambda}} \)  .......(i)
De Broglie wavelength is given by the relation
\(\lambda_{e}=\frac{h}{m v}\)        .........(ii)
On substituting (i) in (ii), we get
\(\lambda_{e}=\frac{h \sqrt{\lambda}}{\sqrt{2 m h c}}=\sqrt{\frac{h \lambda}{2 m c}}\)

(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

T = 27⁰ C = 127 + 273 = 400 K
Energy of neutron at 127⁰ C,
E = 3/2 kT = (3/2) x 1.38 x 10^-23 x  400
= 8.28 x 10^-21 J
Now, \(\lambda =\frac { h }{ \sqrt { 2mE } } =\frac { 6.63\times 10^{ -34 } }{ \sqrt { 2\times 1.66\times 10^{ -27 }\times } 8.28\times 10^{ -21 } }\)
\(=1.264 \ \mathring { A } \)

Kinetic energy of particle of  mass m having momentum p is given by
\(k=\frac { p^{ 2 } }{ 2m } \quad p=\sqrt { 2mK } \)
The de-Broglie wavelength \(\lambda =\frac { h }{ p } =\frac { h }{ \sqrt { 2mK } }\)
\( \\ p=\frac { h }{ \lambda } \)
\(\\ K=\frac { h^{ 2 } }{ 2m\lambda ^{ 2 } } \)
If \(\lambda \) is constant then from Eq (i) we get
p = constant,i.e m_pv_p = m_ev_e
\(\text { or } \frac{v_{p}}{v_{e}}=\frac{m_{e}}{m_{p}}<1\)
or vp < ve
If \(\lambda \) is constant then from (ii)  \(K \propto \frac{1}{m}\)
\((\therefore \frac{K_{p}}{K_{e}}=\frac{m_{e}}{m_{p}}<1 \text { or } K_{p})\)
It means that the velocity of electron is greater than that of proton. Kinetic energy of electron is greater than that of proton.

Number of atoms in 3g coin = \(\frac { 6.023\times { 10 }^{ 23 }\times 3 }{ 63 } =2.868\times { 10 }^{ 22 }\)
Each atom of copper contains 29 protons and 34 neutrons. Therefore, mass defect of each atom
\(=\left[ 29\times 1.00783+34\times 1.00867 \right] -62.92960=0.59225u\)
Total mass defect for all the atoms = \(0.59225\times 2.868\times { 10 }^{ 22 }u\)
\( \Delta m=1.6985\times { 10 }^{ 22 }u\)
As 1u = 931 MeV,
Nuclear energy required = \(1.6985\times { 10 }^{ 22 }\times 931MeV=1.58\times { 10 }^{ 25 }MeV\)

\(\frac { N }{ { N }_{ 0 } } =\frac { 1 }{ 16 } ,\ t=4\quad days,{ N }_{ 0 }=?N=4\times { 10 }^{ -3 }kg.\)
\(As \ \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 1 }{ 2 } \right) }^{ n }=\frac { 1 }{ 16 } ={ \left( \frac { 1 }{ 2 } \right) }^{ 4 },n=4or\frac { t }{ T } =4,\)
\(T=\frac { t }{ 4 } =\frac { 4 }{ 4 } =1 \ day\)
\(Again, \ \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 1 }{ 2 } \right) }^{ n }={ \left( \frac { 1 }{ 2 } \right) }^{ 1/T }={ \left( \frac { 1 }{ 2 } \right) }^{ 6/1 }=\frac { 1 }{ 64 }\)
\({ N }_{ 0 }=64N=64\times 4\times { 10 }^{ -3 }kg=0.256kg\)

(i) 1.1 eV
(ii) \(1.6\times { 10 }^{ -6 }A\)
(iii) unchanged

\(3.77\times { 10 }^{ 21 }\)

\(Here \ { v }_{ 0 }=3V, \ { v }_{ 0 }=6\times { 10 }^{ 14 }{ s }^{ -1 },v=?,{ \phi }_{ 0 }=?\)
\(e{ v }_{ 0 }=hv-h{ v }_{ 0 }\)
\(or \ v={ v }_{ 0 }+\frac { e{ v }_{ 0 } }{ h } =6\times { 10 }^{ 14 }+\frac { { \left( 1.6\times { 10 }^{ -19 } \right) }^{ 3 } }{ 6.6\times { 10 }^{ -34 } }\)
\(=6\times { 10 }^{ 14 }+7.27\times { 10 }^{ 14 }=13.27\times { 10 }^{ 14 }{ s }^{ -1 }\)
\(=\frac { 6.6\times { 10 }^{ -34 }\times 6\times { 10 }^{ 14 } }{ 1.6\times { 10 }^{ -19 } } eV=2.48eV\)

Neutron separation energy \({ S }_{ n }\) of a nucleus \(\\ \begin{matrix} A \\ X \end{matrix}X\) 
Given by \({ S }_{ n }=[{ m }_{ N }(\begin{matrix} A-1 \\ Z \end{matrix}X)+{ m }_{ n }-{ m }_{ N }(\begin{matrix} A \\ X \end{matrix}X)]{ c }^{ 2 }\).
Adding and subtracting the term \({ Zm }_{ C }\) in the bracket above ignoring mass defects due to electronic binding energies we get \({ S }_{ n }\) in terms of atomic masses.

\({ S }_{ n }=[{ m }_{ N }=(\begin{matrix} A-1 \\ Z \end{matrix}X)+{ m }_{ n }-{ m }(\begin{matrix} A \\ Z \end{matrix}X)]{ c }^{ 2 }\)
\({ S }_{ n }(\begin{matrix} 41 \\ 20 \end{matrix}X)=[39.96259+1.008665-40.962278)]u,\)
\(=[40.971255-40.962278]931.5\)
\(=0.008978 \times 931.5\)
\(=8.363 \ MeV.\)
\(\\ Neutron \ separation \ energy \ for \ \begin{matrix} 27 \\ 13 \end{matrix}Al\)
\({ S }_{ n }(\begin{matrix} 27 \\ 13 \end{matrix}Al)=[{ m }_{ N }(\begin{matrix} 26 \\ 13 \end{matrix}Al)+{ m }_{ n }-{ m }_{ N }(\begin{matrix} 27 \\ 13 \end{matrix}Al)]{ c }^{ 2 }\)
\({ S }_{ n }(\begin{matrix} 27 \\ 13 \end{matrix}Al)=[25.986895+1.0008665-26.981541]u\)
\( =[26.99556-26.981541]931.5\)
\( =0.014019 \times 931.5\)
\(=13.06 \ MeV.\)

The minimum energy required to emit the electron, from the ground state of the atom, is called ionization energy.
\(\therefore \text { Ionization energy, } E_{n}=\frac{m e^{4}}{8 n^{2} \varepsilon_{0}^{2} h^{2}}\)
Therefore, the ionization energy of a particle will become 200 times, the ionization energy required by an electron.

Since, work function is given as,
\( W_0=h v_o \\ W_0 \propto v_0 \)
As work function of metal X is higher than metal Y, so metal X has higher threshold frequency than metal Y.

Using Einstein's photoelectric equation,
eV = hv - \(\phi \)
On differention, we get eΔV = hΔv
or \(h=\frac { e\Delta V }{ \Delta N } =\frac { 1.6\times { 10 }^{ -19 }\times \left( 1.23-0 \right) }{ \left( 8-5 \right) \times { 10 }^{ 14 } } \)
\(=6.56\times { 10 }^{ -34 }J-s\)

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

For a particular frequency of incident radiation, the minimum negative (retarding) potential V_o given to plate A for which the photoelectric current becomes zero, is called cut-off or stopping potential.

\(2.12\times { 10 }^{ -10 }\ m\ and\ 4.47\times { 10 }^{ -10 }\ m\)

Metal A has higher value of work function because the slopes of both materials are constant and the intercept of the line depends on the work function.

The photoelectrons can be emitted from a metal surface if the frequency of incident radiation is more than the threshold frequency, i.e., more than that of green light for the given surface. As the frequency of violet light is more than that of green light, hence violet light will eject photoelectrons. But the frequency of red light is less than that of the green light, hence red light can not eject photoelectrons from the given surface.

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.

The energy of blue light \((h v)_{b l u e}\) is greater than the energy of red light. \((h v)_{r e d}\) In photoelectric emission, max K.E. of emitted electron
\(=h v-\phi_0, \text { i.e., } \max K . E \alpha(h v)\)
So, K.E of emitted electrons is more with blue light than that of red light. 

(i) Distinction between Nuclear Fission and Nuclear Fusion Fission is the splitting of large nucleus into two or more smaller ones, on the other hand, fusion is the combining of two or more lighter nuclei to form larger one.
(ii) The given reaction for decay process,
\({ }_{26}^{56} \mathrm{Fe} \longrightarrow 2_{13}^{28} \mathrm{Al}\)
Mass defect,
\(\Delta\)m = m \(\left({ }_{26}^{56} \mathrm{Fe}\right)-2 m\left({ }_{13}^{28} \mathrm{Al}\right)\)
= 55.93494- 2(27.98191)
= - 0.02888
Q = \(\Delta\)m x 931 MeV
= - 0.02888 x 931 MeV
= - 26.88728 MeV

For a dynamically stable orbit in a hydrogen atom,
\( F_{e}=F_{c} \)
\(\frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r_{n}^{2}} =\frac{m v_{n}^{2}}{r_{n}} \)
\(\Rightarrow \quad v_{n} =\frac{e}{\sqrt{4 \pi \varepsilon_{0} r_{n} m}} \)      .......(i)
According to the Bohr's second postulate of quantisation,
\(m v_{n} r_{n}=\frac{n h}{2 \pi}\)      .........(ii)
Combining equations (i) and (ii), we get
\( r_{n} =\left(\frac{n^{2}}{m}\right)\left(\frac{h}{2 \pi}\right)^{2} \frac{4 \pi \varepsilon_{0}}{e^{2}} \)
\(=\frac{\varepsilon_{0} n^{2} h^{2}}{\pi m e^{2}} \)      ...........(iii)
\( \because \text { P.E. } =-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{e^{2}}{r_{n}}=-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{e^{2}}{\frac{\varepsilon_{0} n^{2} h^{2}}{\pi m e^{2}}} \)
\(=-\frac{m e^{4}}{4 \varepsilon_{0}^{2} n^{2} h^{2}} ; \text { K.E. }=\frac{1}{2} m v_{n}^{2}=\frac{m e^{4}}{8 \varepsilon_{0}^{2} n^{2} h^{2}} \)
Total energy of an electron in the stationary states of hydrogen atom is
\(E_{n}=\text { P.E. }+\mathrm{K} . \mathrm{E} .=\frac{-m e^{4}}{4 \varepsilon_{0}^{2} n^{2} h^{2}}+\frac{m e^{4}}{8 \varepsilon_{0}^{2} n^{2} h^{2}}=\frac{-m e^{4}}{8 \varepsilon_{0}^{2} n^{2} h^{2}} .. ..(iv)\)
According to the third postulate of Bohr's model, \(h v_{i f}=E_{n_{i}}-E_{n_{f}}\)
Using equation above, we get \(h v_{i f}=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}}\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)\)
The frequency of radiation emitted is given by \(v_{i f}=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{3}}\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)\)
Now higher energy state, n_i = 3, 2, 1
Lower energy state, n_f = 4
For the transition, n_i = 4 to n_f = 3 \(\rightarrow\) Paschen series
n_i = 4 to n_f = 2 \(\rightarrow\) Balmer series
n_i = 4 to n_f = 1 \(\rightarrow\) Lyman series

(i) This is because the energy of visible photons is less than work function of most of the metals.
(ii) \(\frac { 1 }{ 2 } { { mv }^{ 2 } }_{ max }=hv-{ { \phi }_{ 0 }=hv-{ hv }_{ 0 } }\)
 \({ v }_{ 0 }=v-\frac { { { mv }^{ 2 } }_{ max } }{ 2h } =7.21\times { 10 }^{ 14 }-\frac { \left( 9.1\times { 10 }^{ -31 }kg \right) \times { \left( 6\times { 10 }^{ 5 } \right) }^{ 2 } }{ 2\times \left( 6.63\times { 10 }^{ -34 } \right) } =4.74\times { 10 }^{ 14 }Hz\)
(iii) We find that KE of electron ejected can be increased by increasing the incident energy or by decreasing the work function. In day-to-day life we can prosper by increasing our sincere efforts or by decreasing our requirements.

(a)Potential difference across the evacuated tube, V = 500 V
Specific charge of an electron, e/m = 1.76 x 10¹¹ C kg⁻¹
The speed of each emitted electron is given by the relation for kinetic energy as:
\(K E=\frac{1}{2} m v^{2}=e V\)
\(\therefore v=\left(\frac{2 e V}{m}\right)^{\frac{1}{2}}=\left(2 V \times \frac{e}{m}\right)^{\frac{1}{2}}\)
\(=\left(2 \times 500 \times 1.76 \times 10^{11}\right)^{\frac{1}{2}}=1.327 \times 10^{\mathrm{m} / \mathrm{s}}\)
Therefore, the speed of each emitted electron is \(1.327 \times 10^{7} \mathrm{~m} / \mathrm{s}\)
(b)Potential of the anode, V = 10 MV = 10 x 10⁶ V
The speed of each electron is given as:
\(v=\left(2 V \frac{e}{m}\right)^{\frac{1}{2}}\)
\(=\left(2 \times 10^{7} \times 1.76 \times 10^{11}\right)^{\frac{1}{2}}\)
\(=1.88 \times 10^{9} \mathrm{~m} / \mathrm{s}\)
This result is wrong because nothing can move faster than light. In the above formula, the expression (mv²/2) for energy can only be used in the non-relativistic limit, i.e., forv << c.
For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the total energy is given as:
E = mc²
Where,
m = Relativistic mass
\(=m_{0}\left(1-\frac{v^{2}}{c^{2}}\right)^{\frac{1}{2}}\)
m₀ = Mass of the particle at rest
Kinetic energy is given as:
K = mc² − m₀c²

Wavelength of light of a sodium line, λ = 589 nm = 589 x 10⁻⁹ m
Mass of an electron, m_e= 9.1 x 10⁻³¹ kg
Mass of a neutron, m_n= `1.66 x 10⁻²⁷ kg
Planck’s constant, h = 6.6 x 10⁻³⁴ Js
(a) For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation:
\(K=\frac{1}{2} m_{e} v^{2}\)
We have the relation for de Broglie wavelength as:
\(\lambda=\frac{h}{m_{e} v}\)
\(\therefore v^{2}=\frac{h^{2}}{\lambda^{2} m_{e}^{2}}\)
Substituting equation (2) in equation (1), we get the relation:
\(K=\frac{1}{2} \frac{m_{e} h^{2}}{\lambda^{2} m e_{e}^{2}}=\frac{h^{2}}{2 \lambda^{2} m_{e}}\)
\(=\frac{\left(6.6 \times 10^{-34}\right)^{2}}{2 \times\left(589 \times x 10^{-9}\right)^{2} \times 9.1 \times 10^{-31}}\)
\(= 6.9 \times 10^{-25} J\)
\(=\frac{6.9 \times 10^{-25}}{1.6 \times 10^{-19}}=4.31 \times 10^{-6} \mathrm{eV}=4.31 \mu \mathrm{eV}\)
Hence, the kinetic energy of the electron is 6.9 x 10⁻²⁵ J or 4.31 μeV.
(b) Using equation (3), we can write the relation for the kinetic energy of the neutron as:
\(\frac{h^{2}}{2 \lambda^{2} m_{n}}\)
\(=\frac{\left(6.6 \times 10^{-34}\right)^{2}}{2 \times\left(589 \times 10^{-9}\right)^{2} \times 1.66 \times 10^{-27}}\)
\(=3.78 \times 10^{-28} J\)
\(=\frac{3.78 \times 10^{-28}}{1.6 \times 10^{-19}}=2.36 \times 10^{-9} e V=2.36 \neq V\)
Hence, the kinetic energy of the neutron is 3.78 x 10⁻²⁸ J or 2.36 neV.

\(Here,T=4.5\times { 10 }^{ 9 }years\)
\( =4.5\times { 10 }^{ 9 }\times 365\times 24\times 60\times 60s\)
\(=1.42\times { 10 }^{ 17 }s\)
\(As \ number \ of \ atoms \ in \ 238 \ g\)
\(=Avagadro's\ no.=6.023\times { 10 }^{ 23 }\)
\( \therefore \ number \ of \ atoms \ in \ 1g \ of \ sample\)
\( N=\frac { 6.023\times { 10 }^{ 23 } }{ 238 }\)
\(R=-\frac { dN }{ dt } =\lambda N=\frac { 0.693 }{ T } N\)
\( =\frac { 0.693\times 6.023\times { 10 }^{ 23 } }{ 1.42\times { 10 }^{ 17 }\times 238 }\)
\( =1.235\times { 10 }^{ 3 }Bq\)

\(we \ know, \ 1 \ amu=1.66\times { 10 }^{ -27 }kg\)
\(From \ E={ mc }^{ 2 }=(1.66\times { 10 }^{ -27 }){ \left( 3\times { 10 }^{ 8 } \right) }^{ 2 }\)
\( =1.494\times { 10 }^{ -10 }J\)
\(E=\frac { 1.494\times { 10 }^{ -10 } }{ 1.6\times { 10 }^{ -13 } } MeV=933.75MeV\)
\(For \ oxygen_{ 8 }{ { O }^{ 16 } }, \ mass \ defect\)
\(=8{ m }_{ p }+8{ m }_{ n }-{ M }_{ oxy }\)
\(=8\times 1.00727+8\times 1.00866-15.99053\)
\( =0.13691 \ amu\)
\(=0.13691\times 933.75MeV/{ c }^{ 2 }\)
\( =127.8MeV/{ c }^{ 2 }\)

\(Here,\lambda =5000\overset { \circ }{ A } =5\times { 10 }^{ -7 }m,{ \phi }_{ 0 }=1.9eV\)
(i) Energy of photon in eV,
\(E=\frac { hc }{ \lambda } =\frac { \left( 6.63\times { 10 }^{ -34 } \right) \times \left( 3\times { 10 }^{ 8 } \right) }{ \left( 5\times { 10 }^{ -7 } \right) \times \left( 1.6\times { 10 }^{ 19 } \right) }\)
\(=2.4825eV\)
(ii) Kinetic energy of photelectron
\(=\frac { hc }{ \lambda } -{ \phi }_{ 0 }=2.4825-1.9\)       
\( =0.5825eV\)
(iii) If \({ V }_{ 0 }\) is the stopping potential, then \({ eV }_{ 0 }=K.E.\) of emitted photoelectron = 0.5825 eV  or \({ V }_{ 0 }=\frac { 0.5825eV }{ e } =0.5825V\)