P = 50 mW; λ = 640nm = 640 x 10^-9 m
P = 50 x 10^-3W
\(n=\cfrac { hc }{ \lambda } = \frac{6.626 \times10^{-34} \times 3 \times 10^8}{640 \times 10{-9}}=3.106 \times 10^{-19}J\)
\(n=\frac{E}{hv}=\cfrac { 50\times { 10 }^{ -3 } }{ 3.106\times { 10 }^{ -19 } } = 1.61\times 10^{17} s^{-1}\) 
n = 1.61 x 10¹⁷ s^-I

Due to the large mass of a baseball, the de Broglie wavelength (⋋ = h/mv) associated with a moving baseball is very small. Hence, its wave nature is not visible.

The minimum energy needed for an electron to escape from the metal surface is called work function of that metal.
Unit: electron volt (eV).

It is used in radio and television communication systems and also in cellular phones to transmit voice communication in the ultra high frequency band.

The amplitude of the electric field, E₀ = 3 x 10⁴ NC^-1
The amplitude of the magnetic field, B₀ = 2 x 10^-4 T. Therefore, speed of the electromagnetic wave in a medium is
v = \(\frac { 3\times { 10 }^{ 4 } }{ 2\times { 10 }^{ -4 } } \)  = 1.5 x 10⁸ ms^-1.

\(\mathrm{P}  =\frac{\mathrm{n}}{\mathrm{t}} \frac{\mathrm{hc}}{\lambda}  \)
\(\frac{\mathrm{n}}{\mathrm{t}}  =\frac{\mathrm{P} \lambda}{\mathrm{hc}}  \)
\(\frac{\mathrm{n}}{\mathrm{t}}  =\frac{3.8 \times 10^{26} \times 550 \times 10^{-9}}{6.6 \times 10^{-3} \times 3 \times 10^8}=1 \times 10^{-15}\)

\(\mathrm{E}_{\mathrm{p}}  =\frac{\mathrm{hc}}{\lambda_{\mathrm{p}}}  \)
\(\mathrm{E}_{\mathrm{e}}  =\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda_{\mathrm{e}}^2}  \)
\(\frac{\mathrm{hc}}{\lambda_{\mathrm{p}}}  =\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda_{\mathrm{e}}^2}  \)
\(\lambda_{\mathrm{p}}  \propto \lambda_{\mathrm{e}}^{{ }^2}\)

B_o  = 3 x 10^-6T
Amplitude of electric field, E_o = B_oc
E_o = 3 x 10^-6 x 3 x 10⁸ = 900V m ^-1  

Davisson - Germer experiment
(i) The filament F is heated by a low tension (L . T) battery. Electrons are emitted from the hot filament by thermionic emission.
(ii) They are then accelerated due to the potential diference between the filament and the anode aluminum cylinder by a high tension (H.T) battery.
(iii) Electron beam is collimated by using two thin aluminum diaphragms and is allowed to strike a single crystal of Nickel.
(iv) The electrons scattered by Niatoms in diflerent directions are received by the electron detector which measures the intensity of scattered electron beam.
(v) The detector is capable of rotation in the plane of the paper, so that the angle (\(\theta\)) between the incident beam and the scattered beam can be changed at our will.
(vi) The intensity of the scattered electron beam is measured as a function of the angle \(\theta\).

(i) Figure shows the variation of intensity of the scattered electrons with the angle \(\theta\) for the accelerating voltage of 54 V.
(ii) For a given accelerating voltage V, the scattered wave shows a peak or maximum at an angle of 50^o to the incident electron beam.
(iii) This peak in intensity is attributed to the constructive interference of electrons diffracted from various atomic layers of the target material.


(iv) From the known value of interplanar spacing of Nickel, the wavelength of the electron wave has been experimentally calculated as 1.65\(\overset { o }{ A }\).
(v) The wavelength can also be calculated from de Broglie relation for V = 54 V from equation as
\(\lambda =\cfrac { 12.27 }{ \sqrt { V } } \overset { o }{ A } =\cfrac { 12.27 }{ \sqrt { 54 } } \) 
\(\lambda =1.67\overset { o }{ A } \)
(vi) This value agrees very well with the experimentally observed wavelength of 1.65 \(\overset { o }{ A }\). Thus this experiment directly verifies de Broglie's hypothesis of the wave nature of moving particles.

(i) When the spectrum obtained from the Sun is examined, it consists of large number of dark lines . These dark lines in the solar spectrum are called Fraunhofer lines.
(ii) The Absorption spectra for various materials are compared with the Fraunhofer lines in the solar spectrum, which helps in identifying elements present in the Sun's atmosphere.

(i) The experimental set up it consists of two metal electrodes which are made of small spherical metals. These are connected to larger spheres and the ends of them are connected to induction coil with very large number of turns.
(ii) This is to produce very high electromotive force (emf). Since the coil is maintained at very high potential, air between the electrodes gets ionized and spark (spark means discharge of electricity) is produced. 
(iii) This discharge of electricity affects another electrode (ring type - not completely closed) which is kept at far distance. This implies that the energy is transmitted from electrode to the receiver (ring electrode) in the form of waves, known as electromagnetic waves. If the receiver is rotated by 90^o then no spark is observed by the receiver.
(iv) This confirms that electromagnetic waves are transverse waves as predicted by Maxwell. Hertz detected radio waves and also computed the speed of radio waves which is equal to the speed of light (3 x 10⁸ ms^-1).

Laws of photoelectric effect:
(i) For a given surface, the emission of photoelectrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.
(ii) For a given frequency of incident light, the number of photoelectrons emitted is directly proportional to the intensity of the incident light. The saturation current is also directly proportional to the intensity of incident light.
(iii) Maximum kinetic energy of the photoelectrons is independent of the intensity of the incident light.
(iv) Maximum kinetic energy of the photoelectrons from a given metal is directly proportional to the frequency of incident light.
(v) There is no time lag between the incidence of light and the ejection of photoelectrons.

\(v=10^{14} \mathrm{~Hz} ; \mathrm{E}=19.86 \mathrm{~J} \)
\(E=nhv \Rightarrow n=\frac{E}{hv}\)
\(=\frac{19.86}{6.626 \times 10^{-34} \times 10^{14}}=2.99 \times 10^{20} \)
\(\mathrm{n} \simeq 3 \times 10^{20} \)

(i) An electron of mass m is accelerated through a potential difference of V volt. The kinetic energy acquired by the electron is given by
\(\cfrac { 1 }{ 2 } { mv }^{ 2 }=ev\) 
(ii) Therefore, the speed v of the electron is
\(v=\sqrt { \cfrac { 2ev }{ m } } \) 
Hence, the de Broglie wavelength of the matter waves associated with electron is
\(\lambda =\cfrac { h }{ mv } =\cfrac { h }{ \sqrt { 2mev } } \) 
(iii) Substituting the known values in the above equation, we get
\(\lambda =\cfrac { 6.26\times { 10 }^{ -34 } }{ \sqrt { 2V\times 1.6\times { 10 }^{ -19 }\times 9.11\times { 10 }^{ -31 } } } \) 
= \(\cfrac { 12.27\times { 10 }^{ -10 } }{ \sqrt { V } } m\) 
\(\lambda =\cfrac { 12.27 }{ \sqrt { V } } \overset { o }{ A } \)
(iv) Since the kinetic energy of the electron, K = eV, then the de Broglie wavelength associated with electron can be also written as
\(\lambda =\cfrac { h }{ \sqrt { 2mK } } \)