Given mass of nitrogen molecule
= 2 x 14.0076 u
= \(28.0152\times1.67\times{ 10 }^{ -27 }kg\)
\(T=300k\)
Since de-Broglie wavelength.
 \(\lambda =\frac { h }{ \sqrt { 3mkT } } \)
\(\lambda =\frac { 6.62\times{ 10 }^{ -34 } }{ \sqrt { 3\times28.0152\times1.67\times{ 10 }^{ -27 }\times1.38\times{ 10 }^{ -23 }\times300 } } \)
\(=\frac { 6.62\times{ 10 }^{ -34 } }{ 241 } \)
= \(0.0275\times{ 10 }^{ -9 }\)
= 0.028 nm

(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.

\(E=h c / \lambda\)
= 3.3 x 10^-19 J 

\(\theta=90-\phi / 2\)
\(=90-52 / 2=64^{\circ}\)

\(1 / 2 m v^2=\left(m^2 v^2\right) / 2 m=p_2 / 2 m\)
According to De Broglie wave length \(\lambda=h / p\)
\( \lambda_e / \lambda_p=P_p / P_e=\sqrt{ }\left(m_p / m_e\right) \\ m_e<m_p \\ \lambda_e>\lambda_p \)
electrons have greater De broglie wavelength than proton

Photocells are used for reproduction of sound

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

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.

It is the minimum negative potential given to the anode in a photocell for which the photoelectric current becomes zero. If \({ v }_{ 0 }\) is the stopping potential, then maximum K.E. of emitted photoelectron is
\({ \left( K.E \right) }_{ max }={ eV }_{ 0 }=hv-{ \phi }_{ 0 }\)
\(V_{ 0 }=\frac { hv }{ e } -\frac { { \phi }_{ 0 } }{ e } \)
Threshold frequency  It is the minimum frequency of the incident radiation for which just emission of photoelectrons takes place from a metal surface without any K.E. If \(V_{ 0 }\)is the threshold frequency, then using Einstein's photoelectric equation
\(0={ hv }_{ 0 }-{ \phi }_{ 0 }\quad or\quad V_{ 0 }=\frac { { \phi }_{ 0 } }{ h } \)

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.

Yes; because the wavelength of ultraviolet light is less than the threshold wavelength 5200\(A˚\)

(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. ​

Since the electron has been localised within the nucleus (of size about \({ 10 }^{ -14 }\) m)of an atom, so \({ \lambda =10 }^{ -14 }\)
Momentum of electron,
  \(p=\frac { h }{ \lambda } =\frac { 6.63\times { 10 }^{ -34 } }{ { 10 }^{ -14 } } =6.63\times { 10 }^{ -20 }kg{ ms }^{ -1 }\)
The relativistic relation for the energy of an electron is
\(E=\sqrt { { p }^{ 2 }{ c }^{ 2 }+{ { m }_{ 0 } }^{ 2 }{ c }^{ 4 } } \)
Neglecting the rest-mass energy term(i.e., second term), we have
 \(E=pc=(6.63\times { 10 }^{ -20 })\times (3\times { 10 }^{ 8 })J\)
\( =\frac { 6.63\times 3\times { 10 }^{ -12 } }{ 1.6\times { 10 }^{ -13 } } MeV=124.3MeV\)
This energy E(=124.3MeV) is very large as compared to the binding energy which is provided by Coulomb's force within the nucleus. Since the energy of electron inside the nucleus, hence the electron does not reside the nucleus.

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