(a) For the electron:
Mass m = 9.11 x 10^–31 kg, speed v = 5.4 x 10⁶ m/s. Then, momentum
p = m v = 9.11 x 10^–31 (kg) x 5.4 x 10⁶ (m/s)
p = 4.92 x 10^–24 kg m/s
de Broglie wavelength,\(\lambda\) = h/p
\(=\frac{6.63 \times 10^{-34} \mathrm{Js}}{4.92 \times 10^{-24} \mathrm{~kg} \mathrm{~m} / \mathrm{s}}\)
\(\lambda=0.135 \mathrm{nm}\)
(b) For the ball:
Mass m’ = 0.150 kg, speed v’ = 30.0 m/s.
Then momentum p’ = m’ v ’ = 0.150 (kg) x 30.0 (m/s)
p ’= 4.50 kg m/s
de Broglie wavelength \(\lambda\)’ = h/p’.
\(=\frac{6.63 \times 10^{\pm 34} \mathrm{Js}}{4.50 \times \mathrm{kg} \mathrm{m} / \mathrm{s}}\)
\(\lambda^{\prime}=1.47 \times 10^{-34} \mathrm{~m}\)
The de Broglie wavelength of electron is comparable with X-ray wavelengths. However, for the ball it is about 10^–19 times the size of the proton, quite beyond experimental measurement.

(i)De Broglie wavelength of the neutron, λ = 1.40 x 10⁻¹⁰ m
Mass of a neutron, m_n = 1.66 x 10⁻²⁷ kg
Planck’s constant, h = 6.6 x 10⁻³⁴ Js
Kinetic energy (K) and velocity (v) are related as:
\(K=\frac{1}{2} m_{n} v^{2} \ldots(1)\)
De Broglie wavelength (λ) and velocity (v) are related as:
\(\lambda=\frac{h}{m_{n}} v \ldots(2)\)
Using equation (2) in equation (1), we get:
\(K=\frac{1}{2} \frac{m_{n} h^{2}}{\lambda^{2} m_{n}^{2}}=\frac{h^{2}}{2 \lambda^{2} m_{n}}\)
\(=\frac{\left(6.63 \times 10^{-34}\right)^{2}}{2 \times\left(1.40 \times 10^{-10}\right)^{2} \times 1.66 \times 10^{-27}}=6.75 \times 10^{-21} J\)
Hence, the kinetic energy of the neutron is 6.75 x 10⁻²¹ J or 4.219 x 10⁻²¹ eV.
Hence, the kinetic energy of the neutron is 6.75 x 10⁻²¹ J or 4.219 x 10⁻² eV.
(ii) Temperature of the neutron, T = 300 K
Boltzmann constant, k = 1.38 x 10⁻²³ kg m² s⁻² K⁻¹
Average kinetic energy of the neutron:
\(K \prime=\frac{3}{2} k T\)
\(=\frac{3}{2} \times 1.38 \times 10^{-23} \times 300=6.21 \times 10^{-21} J\)
The relation for the de Broglie wavelength is given as:
\(\lambda \prime=\frac{h}{\sqrt{2 K / m_{n}}}\)
Where
mn = 1.66 x 10^-27Kg
h = 6.6 x 10^-34Js
\(K \prime=6.75 \times 10^{-21} J\)
\(\therefore \lambda \prime=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 6.21 \times 10^{-21} \times 1.66 \times 10^{-27}}}=1.46 \times 10^{-10} \mathrm{~m}=0.146 \mathrm{nm}\)
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.

(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. ​

\(Here,{ v }_{ 1 }=4\times { 10 }^{ 15 }Hz,{ v }_{ 2 }=8\times { 10 }^{ 15 }Hz\)
From Einstein's photoelectric equation, we have
\(e{ V }_{ 0 }=hv-{ \phi }_{ 0 }\)
\(e{ V }_{ 01 }=h{ v }_{ 1 }-{ \phi }_{ 0 } \ and \ e{ V }_{ 02 }=h{ v }_{ 2 }-{ \phi }_{ 0 }\)
\( \therefore \ e\left( { V }_{ 02 }-{ V }_{ 01 } \right) =h\left( { v }_{ 2 }-{ v }_{ 1 } \right)\)
\(or \ { V }_{ 02 }-{ V }_{ 01 }=\frac { h }{ e } \left( { v }_{ 2 }-{ v }_{ 1 } \right) \)
\(=\frac { \left( 6.6\times { 10 }^{ -34 } \right) }{ 1.6\times { 10 }^{ -19 } } \left( 8-4 \right) { 10 }^{ 15 }\)
\( =16.5 \ V\)

An X-ray probe has a greater energy than an electron probe for the same wavelength.
Wavelength of light emitted from the probe, λ = 1 Å = 10⁻¹⁰ m
Mass of an electron, m_e = 9.11 x 10⁻³¹ kg
Planck’s constant, h = 6.6 x 10⁻³⁴ Js
Charge on an electron, e = 1.6 x 10⁻¹⁹ C
The kinetic energy of the electron is given as:
\(E=\frac{1}{2} m_{e} v^{2}\)
\(m_{e} v=\sqrt{2 E m_{e}}\)
Where,
v = Velocity of the electron
m_ev = Momentum (p) of the electron
According to the de Broglie principle, the de Broglie wavelength is given as:
\(\lambda=\frac{h}{p}=\frac{h}{m_{e} v}=\frac{h}{\sqrt{2 E m_{e}}} \)
\(\therefore E=\frac{h^{2}}{2 \lambda^{2} m_{e}} \)
\(=\frac{\left(6.6 \times 10^{-34}\right)^{2}}{2 \times\left(10^{-10}\right)^{2} \times 9.11 \times 10^{-31}}=2.39 \times 10^{-17} J \)
\(=\frac{2.39 \times 10^{-17}}{1.6 \times 10^{-19}}=149.375 \mathrm{eV}\)
Energy of a photon, \(E \prime=\frac{\mathrm{hc}}{\lambda e} e V\)
\(=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{10^{-10} \times 1.6 \times 10^{-19}} \\ \)
\(=12.375 \times 10^{3} \mathrm{eV}=12.375 \mathrm{keV}\)
Hence, a photon has a greater energy than an electron for the same wavelength.

(a) Power of the medium wave transmitter, P = 10 kW = 10⁴ W = 10⁴ J/s
Hence, energy emitted by the transmitter per second, E = 10⁴
Wavelength of the radio wave, λ = 500 m
The energy of the wave is given as:
\(E_{1}=\frac{\mathrm{hc}}{\lambda}\)
Where,
h = Planck’s constant = 6.6 x 10⁻³⁴ Js
c = Speed of light = 3 x 10⁸ m/s
\(\therefore E_{1}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{500}=3.96 \times 10^{-28} \mathrm{~J}\)
= 3 x 10 ³¹
The energy (E₁) of a radio photon is very less, but the number of photons (n) emitted per second in a radio wave is very large.
The existence of a minimum quantum of energy can be ignored and the total energy of a radio wave can be treated as being continuous.
(b)Intensity of light perceived by the human eye, I = 10⁻¹⁰ W m⁻²
Area of a pupil, A = 0.4 cm² = 0.4 x 10⁻⁴ m²
Frequency of white light, ν= 6 x 10¹⁴ Hz
The energy emitted by a photon is given as:
E = hν
Where,
h = Planck’s constant = 6.6 x 10⁻³⁴ Js
∴ E = 6.6 x 10⁻³⁴ x 6 x 10¹⁴
= 3.96 x 10⁻¹⁹ J
Let n be the total number of photons falling per second, per unit area of the pupil.
The total energy per unit for n falling photons is given as:
E = n x 3.96 x 10⁻¹⁹ J s⁻¹ m⁻²
The energy per unit area per second is the intensity of light.
∴ E = I
n x 3.96 x 10⁻¹⁹ = 10⁻¹⁰
\(n=\frac{10^{-10}}{3.96 \times 10^{-19}}\)
= 2.52 x 10⁸ m² s⁻¹
The total number of photons entering the pupil per second is given as:
n_A = n x  A
= 2.52 x 10⁸ x 0.4 x 10⁻⁴
= 1.008 x 10⁴ s⁻¹
This number is not as large as the one found in problem (a), but it is large enough for the human eye to never see the individual photons.

\( { E }_{ v }={ \phi }_{ 0 }+{ K }_{ max }\)
\( \Rightarrow { \phi }_{ 0 }=0\)
\( \Rightarrow \ { E }_{ v }={ K }_{ max }\)
\(\Rightarrow { K }_{ max }=\frac { { p }^{ 2 } }{ 2m } ={ E }_{ v }\)
\( \Rightarrow \ p=\sqrt { 2m{ E }_{ v } } \)
Wavelength \(\left( \lambda \right) \) of emitted electrons,
 \(\lambda =\frac { h }{ p } =\frac { h }{ \sqrt { 2m{ E }_{ v } } } \)

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.

Let m and K be the mass and kinetic energy of a particle. Its de-Broglie wavelength is
\(\lambda =\frac { h }{ \sqrt { 2mK } } or{ \lambda }^{ 2 }=\frac { { h }^{ 2 } }{ 2mK } or \ K=\frac { { h }^{ 2 } }{ 2m{ \lambda }^{ 2 } } \)
\( i.e.,K\alpha \frac { 1 }{ m } \)
\(\frac { { K }_{ e } }{ { K }_{ p } } =\frac { { m }_{ e } }{ { m }_{ p } } >1 or { K }_{ e }>{ K }_{ p }\)

A photoelectric cell converts changes in intensity illumination into changes in electric current. The applications of photoelectric cells are 
(i) in burglar alarm
(ii) in fire alarm
(iii) in the reproduction of sound from films in the cinema hall.