\(\lambda_0  =\frac{h c}{\phi}  \)
\(=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.313 \times 1.6 \times 10^{-19}}  \)
\( =\frac{19.8782400}{5.3} \times 10^{-7}  \)
\(\lambda_0  =3.750 \times 10^{-7} \simeq 3750 \stackrel{o}A\)

\(\frac{\mathrm{hc}}{\lambda}  =\phi+\mathrm{K} . \mathrm{E}  .....(1)  \)
\(\frac{2 \mathrm{hc}}{\lambda}  =\phi+3 \mathrm{~K} . \mathrm{E}......(2)\)
multiply eqn. (1) by 3, we get
\(\frac{3 \mathrm{hc}}{\lambda}  =3\phi+3 \mathrm{~K} . \mathrm{E}  .....(3)\)
Subtract eqn. (2) from (3), we get
\(\frac{\mathrm{hc}}{\lambda}  =2\phi \)
\(\phi=\frac{ \mathrm{hc}}{2\lambda}  \)

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

de Broglie wavelength \(\lambda =\frac { P }{ m } \propto \frac { 1 }{ m } \) for the same velocity v.
\(\frac { { \lambda }_{ p } }{ { \lambda }_{ d } } =\frac { { m }_{ d } }{ { m }_{ p } } =\frac { 2{ m }_{ p } }{ { m }_{ p } } =\frac { 2 }{ 1 } \) = 2.1

The stopping potential does not depend on the intensity of incident radiation; so stopping potential will remain unchanged.

According to de Broglie's hypothesis, all matter particles like electrons, protons, neutrons in motion are associated with waves. These waves are called de Broglie waves or matter waves.

Its working depends on the electron emission from a metal cathode due to irradiation of light or other radiations.

\(\lambda_{1}=3310 Å=3310 \times 10^{-10} \mathrm{~m} ; \mathrm{E}_{1}=3 \times 10^{-19} \mathrm{~J} \)
\(\lambda_{2}=5000 Å=5000 \times 10^{-10} \mathrm{~m} ; \mathrm{E}_{2}=0.972 \times 10^{-19} \mathrm{~J} \)
\(\mathrm{E}=\mathrm{E}_{1}-\mathrm{E}_{2}=2.028 \times 10^{-19} J\)
\(\mathrm{hc}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)=\mathrm{E}_{1}-\mathrm{E}_{2} \)
\(\frac{\mathrm{h} \times 3 \times 10^{8}}{10^{-10}}\left(\frac{1}{3310}-\frac{1}{5000}\right)=2.028 \times 10^{-19} \)
\(\mathrm{~h}=\frac{2.028 \times 10^{-19} \times 10^{-10} \times 3310 \times 5000}{3 \times 10^{8} \times 1690}=6.62 \times 10^{-34} \mathrm{Js} \)
\(\phi_{0} =\frac{\mathrm{hc}}{\lambda}-\mathrm{E}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{3310 \times 10^{-10}}-3 \times 10^{-19} \)
\(=(6-3) \times 10^{-19}=3 \times 10^{-19} \mathrm{~J} \)
\(\phi_{0} =3 \times 10^{-19} \mathrm{~J} \)
Threshold Wavelength,
\(\lambda_{0}=\frac{\mathrm{hc}}{\phi_{0}}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{3 \times 10^{-19}}=6.62 \times 10^{-7} \mathrm{~m} \)
\(\lambda_{0}=6620 \stackrel {o}{A}\)

V = 81 V
∴ K = eV = 1.6 x 10^-19 x 81 = 1.296 x 10^-17 = 1.3 x 10^-17 J
\({ V }=\sqrt \frac {2K}{m}\sqrt { \cfrac { 2\times 1.3\times { 10 }^{ -17 } }{ 9.1\times { 10 }^{ -31 } } } =5.345 \times 10^6 ms^{-1} \) 
v = 5.345 x 10⁶ m s^-1, K = 1.3 x 10^-17 J

de Broglie wavelength \(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}\)
Kinetic energy of electron K \(=\frac{1}{2} m v^{2} \text { (or) } v=\sqrt{\frac{2 K}{m}}\)
Now de Broglie wavelength is \(\lambda=\frac{h}{m v}=\frac{h}{m \sqrt{2 K / m}} \)
\(\lambda=\frac{h}{\sqrt{2 m K}} \)

For a given metallic Surface, the emission of photo electrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.

According to quantum concept, intensity of light of given wavelength is defined as the number of energy quanta or photons incident per unit area per unit time, with each photon having same energy. Its unit is Wm^-2.

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

In metals, the electrons in the outer most shells are loosely bound to the nucleus. Even at room temperature, these large number of electrons which are moving inside the metal in random manner and they cannot leave the surface of metal. So that metals have a large number of free electrons.

(a) λ_e =\(\frac { h }{ { p }_{ e } } \) and λ_p=\(\frac { h }{ { p }_{ p } } \), λ_e = λ_p =1.00 nm.
So, \(\frac { { \lambda }_{ e } }{ { \lambda }_{ p } } =\frac { { p }_{ p } }{ { p }_{ e } } =\frac { 1 }{ 1 } \Rightarrow \frac { { p }_{ p } }{ { p }_{ e } } =\frac { 1 }{ 1 } \)  = 1:1
(b) From relation K=\(\frac { 1 }{ 2 } mv^{ 2 }=\frac { { p }^{ 2 } }{ 2m } \)
K_e = \(\\ \frac { { p }_{ e }^{ 2 } }{ 2me } \) and K_p = \(\frac { { p }_{ e }^{ 2 } }{ 2m_{ p } } \)
\(\frac { { K }_{ p } }{ { K }_{ e } } =\frac { { p }_{ p }^{ 2 } }{ 2{ m }_{ p } } \times \frac { 2{ m }_{ e } }{ { p }_{ e }^{ 2 } } =\frac { { m }_{ e } }{ { m }_{ p } } \)
Since m_e <<< m_p, So K_p <<< K_e
\(\frac { { K }_{ p } }{ { K }_{ e } } =\frac { 9.1\times { 10 }^{ -31 } }{ 1.67\times 10^{ -27 } } \)
= 5.4 x 10^-4

E_k = eV₀ ⇒  6 eV = eV₀ ⇒ V₀ = 6 V
The stopping potential V₀ = 6 volt (Negative).

K.E.= \(\frac{1}{2}mv^2=qV or\)
\(v=\sqrt { \frac { 2qV }{ m } } \)
de Broglie wavelength, \(\lambda =\frac { h }{ mv } \)
\(=\frac { h }{ m\sqrt { \frac { 2qV }{ m } } } =\frac { h }{ \sqrt { 2mqV } } \)
For the same potential difference
\(\frac { { \lambda }_{ \alpha } }{ { \lambda }_{ p } } =\sqrt { \frac { { m }_{ p }{ q }_{ p } }{ { m }_{ \alpha }{ q }_{ \alpha } } } =\sqrt { \frac { { m }_{ p }e }{ { 4m }_{ p }2e } } \)
\(=\frac { 1 }{ 2\sqrt { 2 } } \)

Energy differnce = Energy of emitted photon
= E₂ - E₁
= -1.51 - (-3.4) = 1.89 eV
= 1.89 x 1.6 x 10^-19 J
λ = \(\\ \frac { hc }{ { E }_{ 2 }-{ E }_{ 1 } } \)
= \(\frac { 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 1.89\times 1.6\times 10^{ -19 } } =\frac { 19.8 }{ 3.024 } \) x 10^-7
= 6.548 x 10^-7 m = 6548 Å
The wavelength belongs to the Balmer series of the hydrogen spectrum.

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.

Hertz observation: 
(i) Maxwell's theory of electromagnetism predicted the existence of electromagnetic waves and concluded that light itself is just an electromagnetic wave. Then, the experimentalists tried to generate and detect electromagnetic waves through various experiments.
(ii) In 1887, Heinrich Hertz first became successful in generating and detecting electromagnetic wave with his high voltage spark discharge between two metallic spheres.
(iii) When a spark is formed, the charges will oscillate back and forth rapidly and the electromagnetic waves are produced.
(iv) The electromagnetic waves thus produced were detected by a detector that has a copper wire bent in the shape of a circle.
(v) Although the detection of waves is successful, there is a problem in observing the tiny spark produced in the detector.
(vi) In order to improve the visibility of the spark, Hertz made many attempts and finally noticed an important thing that small detector spark became more vigorous when it was exposed to ultraviolet light.
(vii) The reason for this behavior of the spark was not known at that time. Later it was found that it is due to the photoelectric emission. whenever ultraviolet light is incident on the metallic sphere, the electrons on the outer surface are emitted which caused the spark to be more vigorous.
Hallwachs' observation:
(i) In 1888, Wilhelm Hallwachs, a German physicist, confirmed that the strange behaviour of the spark is due to the action of ultraviolet light with his simple experiment.
(ii) A clean circular plate of zinc is mounted on an insulating stand and is attached to a gold leaf electroscope by a wire.
(iii) When the uncharged zinc plate is irradiated by ultraviolet light from an arc lamp, it becomes positively charged and the leaves will open.
(iv) Further, if the negatively charged zinc plate is exposed to ultraviolet light, the leaves will close as the charges leaked away quickly.
(v) If the plate is positively charged, it becomes more positive upon UV rays irradiation and the leaves will open further.
(vi) From these observations, it was concluded that negatively charged electrons were emitted from the zinc plate under the action of ultraviolet light.

Lenard's observation:
(i) ln 1902, Lenard studied this electron emission phenomenon in detail. His simple experimental setup is as shown in Figure.
(ii) The apparatus consists of two metallic plates A and C placed in an evacuated quartz bulb. The galvanometer G and battery B are connected in the circuit.


(iii) When ultraviolet light is incident on the negative plate C, an electric current flows in the circuit that is indicated by the deflection in the galvanometer.
(iv) On other hand, if the positive plate is irradiated by the ultraviolet light, no current is observed in the circuit.
(v) From these observations, it is concluded that when ultraviolet light falls on the negative plate, electrons are ejected from it which are attracted by the positive plate A.
(vi) On reaching the positive plate through the evacuated bulb, the circuit is completed and the current flows in it.
(vii) Thus, the ultraviolet light falling on the negative plate causes the electron emission from the surface of the plate.

(i) The photoemission of electrons does not depend on the intensity but it depends on the frequency and hence on the energy of a photon of incident light.
(ii) If the energy of a photon is greater than the work function, the photoemission of electrons results however weak the incident radiation may be.
(iii) The energy of a photon of red light is less than the work function of zinc, so red light cannot emit photoelectrons.
(iv) The energy of a photon of ultraviolet light is greater than the work function of zinc, so ultraviolet light can emit photoelectrons.

1) The frequency of incident radiation was kept constant.
2) de Broglie wavelength,
\(\lambda =\frac { h }{ \sqrt { 2mqV } } \alpha \frac { 1 }{ V } \)
If potential difference V is doubled, the de-Broglie wavelength is decreased to \(\frac { 1 }{ \sqrt { 2 } } \) time.

An electron has a larger wavelength.
Reason: de- Broglie wavelength in terms of kinetic energy \(\lambda =\frac { h }{ \sqrt { 2m{ E }_{ K } } } \alpha \frac { 1 }{ \sqrt { m } } \) is for the same kinetic energy.
As an electron has a smaller mass than a proton, an electron has a larger de Broglie wavelength than a proton for the same kinetic energy.

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.

a) The number of photons incident normally per unit area per unit time is determined by the intensity of radiations.
b) (i) Red light, because the energy of red light is less than that of blue light
(hv)_R < (hv)_B
(ii) Blue light, because the energy of blue light is greater than that of red light
(hv)_B > (hv)_R

i) The work function is given by
ϕ₀ = hv - K_max = \(\frac { hc }{ \lambda } \) - eV₀
since K_max = eV₀
\(=\left[ \frac { 6.626\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 390\times 10^{ -9 } } \right] \) - [1.6 x 10^-19 x 1.10]
= 5.10 x 10^-19 - 1.76 x 10^-19 = 3.34 x 10^-19 J
= 2.09 eV
ii) The threshold wavelength is
\(\lambda_{0}=\frac{h c}{\phi_o}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{3.34 \times 10^{-19}}\) 
= 5.951 x 10^-7 m = 5951 \(\mathring { A }\).

\(\text { (i) } \lambda_{\mathrm{d}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV_1}}} ; \lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{\mathrm{me}}}\)
\(\frac{\lambda_{\mathrm{d}}}{\lambda_{\alpha}}=\frac{\frac{1}{\sqrt{2 m e}}}{\frac{1}{\sqrt{4 m 2 e}}} \Rightarrow \lambda_{\mathrm{d}}=2 \lambda_{\alpha}\)
\(\text { (ii) } =\frac{K.E_d}{K.E_a}=\frac{eV}{2eV} =\frac{1}{2}\Rightarrow K.E_d=\frac{1}{2}K.E_a\)
∴ K.E of deuteron is half of K.E of α-particle.

\(\text { K.E }=81.9 \times 10^{-15} \mathrm{~J} \)
\(\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}=\frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-3} \times 1836 \times 81.9 \times 10^{-15}}} \)
\(\lambda =\mathbf{4 . 0 0} \times 10^{-14} \mathrm{~m} \)

\(\lambda=6000Å=6000 \times 10^{-10} \mathrm{~m} ; \mathrm{V}=0.8 \mathrm{v} \)
\(\mathrm{k} \cdot \mathrm{E}=\mathrm{hv}-\phi \)
\(\mathrm{eV}_o=\mathrm{hv}-\phi=\frac{\mathrm{hc}}{\lambda}-\phi \)
\((i) v=\frac{c}{\lambda}=\frac{3 \times 10^{8}}{6000 \times 10^{-10}}=5 \times 10^{14} \mathrm{~Hz} \)
\((ii)\ \mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{6000 \times 10^{-10}}=3.313 \times 10^{-19} J\)
\(\mathrm{E}=\frac{3.313 \times 10^{-19}}{1.6 \times 10^{-19}}=2.07 \mathrm{eV} \)
\((iii) \ \mathrm{E}=\mathrm{hv}-\mathrm{W}\Rightarrow\mathrm{W}=\mathrm{h} v-\mathrm{E} \)
\(\mathrm{E}=\mathrm{eV_o}=1.6 \times 10^{-19} \times 0.8=1.2 8 \times10^{-19}J\)
\(\mathrm{hv}=6.626 \times 10^{-34} \times 5 \times 10^{14}=3.313 \times 10^{-19}J \)
\(\mathrm{~W}=\frac{(3.313-1.28) \times 10^{-19}}{1.6 \times 10^{-19}}=1.270 \mathrm{eV} \)
W = 1.27 eV
\((iv) \ \mathrm{W}=\mathrm{h} \mathrm{v}_{0} \)
\(v_{0}=\frac{W}{h}=\frac{2.033 \times 10^{-19}}{6.626 \times 10^{-34}}=3.07 \times 10^{14} \mathrm{~Hz} \)
\((v)\ \mathrm{E}=\mathrm{eV}_o=\frac{0.8 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}=\mathbf{0 . 8} \mathrm{eV}\)

(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
\(\lambda =\cfrac { h }{ \sqrt { 2mK } } \)