Particle Nature of Light

Photons $h\nu$

Metal

$h \nu=\phi+k \cdot \varepsilon .$

Einstein's Photo electric Equation

Crystal

$2 d \sin \theta = n \lambda$

$\lambda = \frac{h}{\sqrt{2 ~m e v}}$

Max Plank (1900): First breakthrough when Max Plank introduce the concept of Quantum of energy. The black body was explained in the way that energy exchanged between radiation and matter takes place in discrete/quantized way

Einstein Photoelectric effect (1905, Nobel prize in 1921): Quantization of light in the form of photons

Bohr model (1911): Combining the quantization of energy exchange and between the levels of atom

Compton (1923): Confirm the particle nature of the light

De Brogli (1923): The wave nature of particle. So particle behave like wave

Davission and Germer (1927): Interference pattern of electron beam from crystal

Take any two Values.

$\frac{h c}{\lambda} =\phi+K E$

$k E =\frac{h c}{\lambda}-\phi\rightarrow(1)$

• $2.0 \times 1.6 \times 10^{-19} =\frac{h \times 3 \times 10^8}{0.3 \times 10^{-6}(\mathrm{~m})}-\phi \rightarrow(2)$

$1.0 \times 1.6 \times 10^{-19} =\frac{h \times 3.0 \times 10^8}{0.4 \times 10^{-6}(\mathrm{~m})}-\phi\rightarrow (3)$

$(2) -(3)$

$1.6 \times 10^{-19} =\frac{h \times 3.0 \times 10^8}{10^{-7}}(\frac{1}{3}-\frac{1}{4})=\frac{h \times 3.0 \times 10^8}{10^{-7}}\left(\frac{1}{12}\right)$

$h =\frac{12 \times 10^{-7} \times 10^{-19} \times 1.6}{3.0 \times 10^8}$

$h =6.4 \times 10^{-34} \mathrm{~J} \cdot \mathrm{sec}$

The electric field at a point associated with a light wave is,$E=(100 Vm^{-1})$

$\sin[(3.0\times 10^{15} s^{-1})t].$

$\sin[(6.0 \times 10^{15} s^{-1})t].$

If this light falls on a metal surface having a work function of $2.0 \mathrm{eV}$, what will be the maximum kinetic energy of the photoelectrons?

The electric field associated with in rident light:

$E=100$

$\sin [(3.0 \times 10^{15} \mathrm{s}^{-1}) t]$.

$\sin [(6.0 \times 10^{15} \mathrm{s}^{-1}) t]$ = 100 $\times \frac{1}{2} \cos \left(9.0 \times 10^{15} t\right) -\cos \left(3 \times 10^{15} t\right)$

$[2 \sin A \sin B=\cos (A+B)-\cos (A-B)]$

So $\omega_1=9 \times 10^{15} / \omega_2=3 \times 10^{15}$

For $\max k \cdot (...),$

$\omega_1=9 \times 10^{15}$

$\text { So } f_{\text {max }} =\frac{\omega}{2 \pi}$

$f_{\text {max }} =\frac{9 \times 10^{15}}{2 \times 3.14}$

$h \nu=k. E. +\phi$

$K .E=h \nu-\phi = \frac{6.63 \times 10^{-34} \times 9.0 \times 10^{15}}{2 \times 3.14 \times 1.6 \times 10^{-19}}ev-2.0 \mathrm{eV}=(5.93-2) ev=3.93 eV$

$k .E_{\max }=3.93 \mathrm{eV}$

$\frac{h c}{\lambda}=k \cdot \varepsilon+\phi$

$k \cdot \varepsilon=\frac{p^2}{2 m}(p- \text {linear momentum })$

$\frac{p^2}{2 m}=\frac{h c}{\lambda}-\phi\rightarrow (1)$

$\frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{400 \times 10^{-9}}-2.5 \times 1.6 \times 10^{-19}=4.97 \times 10^{-19}-4.0 \times 10^{-19}$

$\frac{p^2}{2 m}=0.97 \times 10^{-19}$

$p =\sqrt{0.97 \times 10^{-19} \times 2 \times 9.1 \times 10^{-31}}$

$p =4.2 \times 10^{-25} {~kg} \cdot \frac{m}{sec}$

$\lambda =350 \mathrm{~nm}$

$\phi =1.9 e v$

$\text{Max} ~k. \varepsilon. =? ?$

$\frac{h c}{\lambda} =k. \varepsilon. +\phi$

$k. \varepsilon. =\frac{h c}{\lambda}-\phi = \frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{350 \times 10^{-9}}J-1.9 \mathrm{eV}$

$=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{350 \times 10^{-9} \times 1.6 \times 10^{-19}} \mathrm{e V}-1.9 e V$

$k. \varepsilon. =1.65 ~eV$

Energy of incident Photon

$=\frac{5 \times 10^{-3}}{8 \times 10^{15}}=6.25 \times 10^{-19}$

$\frac{h c}{\lambda} =\phi+k \cdot \varepsilon .$

$\phi =\frac{h c}{\lambda}-k \varepsilon=6.25 \times 10^{-19}-2 \times 1.6 \times 10^{-19}$

$\phi =3.05 \times 10^{-19}$

$\phi =\frac{3.05 \times 10^{-19}}{1.6 \times 10^{-19}}=1.906 ~{eV}$

The UV light of wavelength $450 \mathrm{~nm}$ and intensity 2.0 $\mathrm{W} / \mathrm{cm}^2$ was shine on a metal surface. Calculate the amount of current flows in the outer circuit due to photoelectrons emitted from the metal surface having the area of $2 \mathrm{cm}^2$ considering only 5 percentage of incident photons produces photoelectrons. Consider the energy of the photons is more than the work function of the metal and efficiency to collect the photoelectrons is 100 %.

$\Rightarrow$ Incident Wavelength $(\lambda)=450 nm$

$\text{Intensity} (I)=2.0 \mathrm{~W} / \mathrm{cm}^2$

$=2.0 \mathrm{~J} / \mathrm{cm}^2 \cdot \mathrm{sec}$

$\text { No. of Photoelectron }$

$=45.24 \times 10^{17} \times \frac{5}{100}=2.263 \times 10^{17}$

$\text{Current flew in outer circuit}$

$=n \times e$

$=2.263 \times 10^{17} \times 1.6 \times 10^{-19}$

$=36 \mathrm{~mA}$

$k.\varepsilon_1 = 0.5 ~e V$

$k.\varepsilon_2 = 1.2 ~e V$

$\frac{h c}{\lambda_1}=\phi+k. \varepsilon_1-(1) \mid k. \varepsilon.=0.5 \mathrm{eV}$

$\frac{h c}{\lambda_2}=\phi+k. \varepsilon_2-(2) \mid k. \varepsilon.=1.2 \mathrm{eV}$

$\phi=$ Work Function and it is the Material property

Using Equation-1

$\frac{h c}{532 \mathrm{~nm}}=\phi+0.5 \mathrm{ev}$

$= \frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{532 \times 10^{-6 \mathrm{~m}}}$

=$\phi+0.5 \times 1.6 \times 10^{-19} = 3.7 \times 10^{-19}$

$=\phi+0.8 \times 10^{-19}$

$\phi=2.9 \times 10^{-19}$

Using Equation-2

$\frac{h c}{\lambda_ 2}=\phi+1.2 \mathrm{ev} =\frac{6.63 \times 10^{-34} \times 3.0 \times 10^8} {\lambda_2(\mathrm{~m})}=2.9 \times 10^{-19}+1.2 \times 1.6 \times 10^{-19}$

$\lambda_2=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{4.82 \times 10^{-19}}$

$\lambda_2=4.12 \times 10^{-7} \mathrm{~m}$

$\lambda_2=412 \mathrm{~nm}$

$h\nu=\phi+K E .$

$\frac{h c}{\lambda_1}=\phi+k \cdot E_1-(1)$

$\frac{h c}{\lambda_2}=\phi+K \cdot E_2-(2)$

$k \cdot E_1=\frac{1}{2} m v_1^2-(3)$

$k \cdot E_2=\frac{1}{2} m v_2^2-(4)$

$\frac{h c}{\lambda_1}-\phi=\frac{1}{2} m v_1^2-(5)$

$\frac{h c}{\lambda_2}-\phi=\frac{1}{2} m v_2^2-(6)$

Divide (5/6)

$\frac{v_1^2}{v_2{ }^2}=\frac{\frac{h c}{\lambda_1}-\phi}{\frac{h c}{\lambda_2}-\phi}-(7)$

$\left(\frac{3}{1}\right)^2 =9=\frac{\frac{h c}{\lambda_ 1}-\phi}{\frac{h c}{\lambda_2}-\phi}$

$\Rightarrow 8 \phi =\frac{9 h c}{\lambda_ 2}-\frac{h c}{\lambda_1}$

$8 \phi =\frac{9 \times 6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{310\times{10^{-9}}}-\frac{663 \times 10^{-34} \times 3.0 \times 10^{8}}{24 \times 10^{-9}}$

$8 \phi =0.497 \times 10^{-17} / 1.6 \times 10^{-19}$

$\Rightarrow \phi=3.88 \mathrm{ev}$

$\phi=4.5 \mathrm{e v}$

$\lambda=200 \mathrm{~nm}$

$\frac{h c}{\lambda}=k .E+\phi$

$K. E=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{200 \times 10^{-9}}-4.5 \times 1.6 \times 10^{-19}$

$=9.945 \times 10^{-19}-7.2 \times 10^{-19}$

$k. E.=2.745 \times 10^{-19}$

$K . E .=\frac{2.745 \times 10^{-19}}{1.6 \times 10^{-19}}=1.7 \mathrm{eV}$

So electron emitting from emitter with $1.7 \mathrm{eV}$ energy

So min entry will be $\simeq 2 \mathrm{eV}$ (Equal to voltage applied)

$\text { Max energy }=(2.0+1.7) \mathrm{eV}=3.7 \mathrm{eV})$

$\text{V = 1.1 Volt}$

$\lambda=400 \mathrm{nm}$

$\text { and } k \cdot E=1.1 \mathrm{ev}$

$\frac{h c}{\lambda}=k \cdot E + \phi$

$\frac{h c}{\lambda}=k \cdot E+\frac{h c}{\lambda_ 0}\left(\phi=\frac{h c}{\lambda_0}\right)$

$\frac{h c}{\lambda_0}=\frac{h c}{\lambda}-k \cdot E \cdot-(1)$

$\frac{h c}{\lambda_0} =\frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{400 \times 10^{-9} \mathrm{m}}-1.1 \times 1.6 \times 10^{-19}$

$=4.97 \times 10^{-19}-1.76 \times 10^{-19}$

$=3.21 \times 10^{-19}$

$\lambda_0 =\frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{3.21 \times 10^{-19}}$

$=6.20 \times 10^{-7} \mathrm{m}$

Wavelength of incident light $=450 \mathrm{~nm}$

So $\frac{h c}{\lambda}$=$k \cdot E.+\phi-(1)$

$\phi=2.0 \mathrm{eV}$

$=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{450 \times 10^{-9} \mathrm{~m}}$

$=\frac{1}{2} m v^2+2.0 \times 1.6 \times 10^{-19}$

$\Rightarrow \frac{1}{2} m v^2=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{450 \times 10^{-9}}-2.0 \times 1.6 \times 10^{-19}$

$=1.22 \times 10^{-19} \quad\left[m v=\sqrt{2 \times m \times 1.22 \times 10^{-19}}\right]$

$m v=\sqrt{2 \times 9.1 \times 10^{-31} \times 1.22 \times 10^{-19}}$

$m v=4.67 \times 10^{-25} \mathrm{~kg} \frac{m}{\text { sec }}-(1)$

Now electron sent $\perp$ to the magnetic field:

$r =\frac{m v}{q B}$

$B =\frac{m v}{q \times r}=\frac{4.67 \times 10^{-25}}{1.6 \times 10^{-19} \times 0.2}$

$B =1.46 \times 10^{-5} \mathrm{~T}$

The electric field associated with a light wave is given by:

$E=E_0 \sin \left[\left(1.57 \times 10^7 \mathrm{~m}^{-1}\right)(x-c t)\right].$

Find the stopping potential when this light is used in an experiment on photoelectric effect with the emitter having work function $1.9 \mathrm{eV}$

$w =1.57 \times 10^7 \mathrm{C}$

$f =\frac{1.57 \times 10^7 \times 3.0 \times 10^8}{2 \times 3.14}\left(\frac{w}{2 \pi}\right) \cdot \mathrm{Hz}$

$\phi=1.9 \mathrm{eV}$

so

$h \nu= K \cdot E.+\phi$

$(K . E .)_{e l e} =h \nu-\phi$

$=\frac{6.63 \times 10^{-34} \times 1.57 \times 10^7 \times 3.0 \times 10^8}{2 \times 3.14 \times 1.6 \times 10^{-19}} \mathrm{eV}-1.9$

$=(3.107-1.9) \mathrm{eV}$

$(K . E.)_{\text {electron }}=1.207 \mathrm{eV}$

So stopping potential need to stop electron

$V=1.207 \mathrm{~Volt}$

Fringe width

$y=1.0 \mathrm{mm} \times 2=2. \mathrm{0mm}$

$d=0.24 \mathrm{mm}$

$\phi=2.2 \mathrm{eV}$

$D=1.2 \mathrm{m}$

$y=\frac{\lambda D}{d}$

$\lambda=\frac{y d}{D}$

$\lambda=\frac{2 \times 10^{-3} \times 0.24 \times 10^{-3}}{1.2}$

$\lambda=4 \times 10^{-7} \mathrm{~m}$

$E= \frac{h c}{\lambda}=\frac{6.63 \times 10^{-34} \times 30 \times 10^8}{4 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{eV}$

$E=3.105 \mathrm{eV}$

Stopping potential:-

$e V_0=3.105-2.2 =0.905 \mathrm{eV}$

$V_0=\frac{0.905 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}=0.905 \mathrm{~V}$

Charge density

$(\sigma)=1 \times 10^{-9} \frac{\mathrm{C}}{\mathrm{m}^2}$

$\phi=1.9 \mathrm{ev}$

$\lambda=400 \mathrm{nm}$

$d=20 \mathrm{cm}=0.2 \mathrm{m}$

Electric Potential due Charge plate

$V=E \times d$

Electric field:

$E=\frac{\sigma}{\epsilon_0}$

So $V =\frac{\sigma}{\epsilon_0} \times d=\frac{1 \times 10^{-9} \times 20}{8.85 \times 10^{-12} \times 100}$

$V =22.7 \mathrm{~V}$

$\Rightarrow \frac{h c}{\lambda} =\phi+k \cdot E .$

$K E =\frac{h c}{\lambda}-\phi-(1)$

$\frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{400 \times 10^{-9} \times 1.6 \times 10^{-19}}-1 . 9 e v$

$=3.105-19$

$K. E. =1.205$

This $K . E.< < e V$ between the plates

So min K. E. is $22.7 \mathrm{eV}$ and max. K.E. $=(22.7+1.205) \mathrm{eV}$

$K. E._{\text {max }}=23.905 \mathrm{eV}$

$K. E._{\text {min }}=22.7 \mathrm{e V}$

$\lambda =400 \mathrm{~nm}$

$\phi =2.2 \mathrm{eV}$

$E =\frac{h C}{\lambda}=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{400 \times 10^{-9} \times 1.6 \times 10^{-19}}=3.1 \mathrm{eV}$

After Collision, Energy lost.

$\begin{aligned}& 3.1 \mathrm{ev} \times \frac{10}{100} \& =0.31 \mathrm{ eV}\end{aligned}$

So energy remains after $1^{\text {St }}$ Collision

$3.1 \mathrm{ev}-0.31 \mathrm{ev}=2.7 \mathrm{9ev}$

After $2^{\text {nd }}$ Collision, energy lost

$\frac{2.79}{100} \times 10=0.279$

So energy remains after $2^{\text {nd }}$ collision

$\begin{aligned}& 2.79-0.279 \& =2.511 \mathrm{eV}\end{aligned}$

After $3^{\text {rd }}$ Collision, energy lost

$\frac{2.511}{100} \times 10=0.2511 \mathrm{eV}$

So energy remains after $3^{\text {rd }}$ Collision

$\begin{array}{r}(2.511-0.2511) \mathrm{eV} \=2.2599 \mathrm{eV}\end{array}$

Energy lost after $4^{Th}$ Collision

$=\frac{2.2599}{100} \times 10 =0.22599 \mathrm{eV}$

So energy remains after $4^{Th}$ Collision

$=2.2599-0.22599 \mathrm{eV}=2.033 \mathrm{eV}$

Therefore, energy of electron after $4^{Th}$ Collision is $2.033 \mathrm{eV}$

$E_{\text {ele }}^{4 ^\text { Th }}<\phi_{\text {metal }}$

So after $4^{\text {Th }}$ Collision electron will not be able to Ceme out.

Energy of Photon

$=\frac{h c}{\lambda}=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{50 \times 10^{-9} \times 1.6 \times 10^{-19}}ev$

$=24.848 \mathrm{~V}$

Energy required to remove $e$ from

$n=1 \text { to } n=\infty = 13.6 \mathrm{eV}$

$\text{So K.E. of ele}$

$=24.84 \mathrm{eV}-13.6 \mathrm{eV}=11.24 \mathrm{eV}$

Radiation in the range of $450 \mathrm{~nm}$ to $550 \mathrm{nm}$ Energy corresponding to $4.50 \mathrm{nm}$

$\frac{1240 \text{ev nm}}{450 \mathrm{nm}}=2.75 \mathrm{eV}$

For $550 \mathrm{~nm}$ :

$\frac{1240 ~{ev. ~nm }}{550 nm} =2.26 \mathrm{eV}$

The light come under visible range so transition from $n=2$

$\text { to } n=3,4,5$

So energy corresponding to these transition

$E_2-E_3 =13.6 \mathrm{eV}\left(\frac{1}{4}-\frac{1}{9}\right)=1.9 \mathrm{eV}$

$E_2-E_4 =13.6 \times\left(\frac{1}{4} - \frac{1}{16}\right) =2.55 \mathrm{eV}$

$E_2-E_5 =13.6 \times\left(\frac{1}{4}-\frac{1}{2}\right)=2.856 \mathrm{eV}$

So only $E_2-E_4$ Comes in the range

So Wavelength Corresponding to this E

$\lambda=\frac{1240 \mathrm{ev} . \mathrm{nm}}{2.55}=486 \mathrm{~nm}$

$\lambda=100 \times 10^{-12} \mathrm{~m}$

$D=40 \mathrm{~cm}$

$\beta=0.1 \mathrm{~mm}$

So distance between two successive maxima

$\beta=\frac{\lambda D}{d}$

$d = \frac{\lambda D}{\beta}= \frac{100 \times 10^{-12} \times 40 \times 10 ^{-2}}{0.1\times 10 ^{-3}}$

$d = 4\times 10 ^{-7} m$

$V=30 \mathrm{kV}$

$i=30 \mathrm{~mA}$

$i=n e$

$n=\frac{i}{e}=\frac{3.0 \times 10^{-3}}{1.6 \times 10^{-19}}=0.625 \times 10^{17}$

$\text { Photoelectrons per sec }$

$k . E.$ of one ele.

$1.6 \times 10^{-19} \times 40 \times 10^3$

$= 6.4 \times 10 ^{-15} J.$

$T_{k\epsilon}= 0.625 \times 6.4\times 10 ^{-15} \times 10 ^{17}= 4.0 \times 10 ^2 J$

$\text{Power emitted X-ray:}$

$4 \times 10 ^2 \times\frac{1}{100} = 4 W$

$V =40 \mathrm{KV}=40 \times 10^3 \mathrm{~V}$

Energy utilized

$\frac{70}{100} \times 40 \times 10^3=28 \times 10^3$

$\lambda=\frac{h c}{E} =\frac{1240 \mathrm{ev} \cdot \mathrm{nm}}{28 \times 10^3 \mathrm{eV}}=44 \mathrm{~pm}$

For other wave length

$E =70 / \text { (left our energy) }=\frac{70}{100}(40-28) \times 10^3=84 \times 10^2$

$\lambda =\frac{h c}{E}=\frac{1240 \mathrm{ev} . \mathrm{nm}}{84 \times 10^2 \mathrm{~V}} =148 \mathrm{pm}$

For third Wave length

E =$\frac{70}{100}(12-8.4) \times 10^3 =25.2 \times 10^2$

$\lambda = \frac{hc}{E}= \frac{1240 {~ev}. nm}{25.2 \times 10^2}= 493 pm$