Modern Physics PDF - Shailendra Pandey Sir

Okay, here is the content of the images converted into a structured markdown format. ### FINAL MARATHON CLASS-12TH Physics **MODERN PHYSICS** **Shailendra Pandey Sir** B.Tech. HBTI, M.Tech. HBTI & M.Sc. Physics --- ### Topics to be covered MODERN PHYSICS --- **SHAILENDRA SIR** JOIN MY OFFI...

Okay, here is the content of the images converted into a structured markdown format. ### FINAL MARATHON CLASS-12TH Physics **MODERN PHYSICS** **Shailendra Pandey Sir** B.Tech. HBTI, M.Tech. HBTI & M.Sc. Physics --- ### Topics to be covered MODERN PHYSICS --- **SHAILENDRA SIR** JOIN MY OFFICIAL TELEGRAM CHANNEL Scan QR code to join. --- ### PHOTOELECTRIC EFFECT The phenomenon of emission of electrons from a metal surface, when electromagnetic radiations of sufficiently high frequency are incident on it, is called photoelectric effect. The photo (light)-generated electrons are called photoelectrons. * Alkali metals like $Li, Na, K, Cs$ and $Rb$ are highly photosensitive. They emit electrons even with visible light. * Metals like $Zn, Cd, Mg, Al$, etc. respond only to ultraviolet light. * $X-rays$ can eject electrons even from heavy metals. --- ### EXPERIMENTAL STUDY OF PHOTOELECTRIC EFFECT 1. Effect of intensity of light on photoelectric current Since the photoelectric current is directly proportional to the number of photoelectrons emitted per second, this implies that the number of photoelectrons emitted per second is proportional to the intensity of incident radiation. *Graphic: A graph that shows Photoelectric current on the y-axis versus Intensity of light on the x-axis. The current increases as the intensity increases.* --- 2\. Effect of potential The value of the retarding potential at which the photoelectric current becomes zero is called cut off or stopping potential for the given frequency of the incident radiation. $K_{max} = \frac{1}{2} m v_{max}^{2} = e V_{0}$ *Graphic: A graph of photocurrent versus anode potential. The point at which retarding potential is equal to zero is marked on the graph.* --- 3\. Effect of frequency of incident radiation on stopping potential The value of stopping potential increases with the frequency of incident radiation. For frequencies $V_{3} > V_{2} > V_{1}$, the corresponding stopping potentials vary in the order $V_{03} > V_{02} > V_{01}$. *Graphic: A graph of photoelectric current vs anode potential is shown with Saturation Current labeled.* --- The stopping potential increases linearly with the frequency $v$ of the incident radiation for a given photosensitive material. For two different metals $A$ and $B$, these graphs are parallel straight lines i.e., they have same slope. But the threshold frequencies are different for the two metals. $h v = W + K_{max}$ $h v = W_{o} + e V_{o}$ $e V_{o} = h v - W_{o}$ $V_{o} = \frac{h}{e} v - \frac{W_{o}}{e}$ $y = mx + c$ *Graphic: A graph of Stopping Potential (Vo) vs Frequency of Incident radiation is shown for Metal A and Metal B. Both lines are shown to be linear and having parallel slope.* --- ### QUESTION Monochromatic light of frequency $6.0 \times 10^{14} Hz$ is produced by a laser. The power emitted is $2.0 \times 10^{-3} W$. (i) What is the energy of each photon in the light? (ii) How many photons per second, on the average, are emitted by the source? \[NCERT] Solution: Here $v = 6.0 \times 10^{14} Hz, P = 2.0 \times 10^{-3} W$ (i) Energy of each photon, $E = h v= 6.63 \times 10^{-34} \times 6.0 \times 10^{14} = 3.98 \times 10^{-19} J$ **(ii)** If $N$ is the number of photons emitted per second by the source, then Power transmitted in the beam $P = N \times$ energy of each photon $P = N E$ $N = \frac{P}{E} = \frac{2.0 \times 10^{-3} W}{3.98 \times 10^{-19} J}$ $N = 5.0 \times 10^{15}$ photons per second. --- ### QUESTION The work function of caesium is 2.14 eV. Find (a) the threshold frequency for caesium, and (b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 eV. \[NCERT] Solution: (a) For the minimum, cut-off or threshold frequency, Energy $hv_{o}$ of incident photon = Work function $W_{o}$ $v_{o} = \frac{W_{o}}{h}$ $v_{o} = \frac{2.14eV}{6.63 \times 10^{-34}Js} = \frac{2.14 \times 1.6 \times 10}{6.63 \times 10^{-34}Js}$ $v_{o} = 5.16 \times 10^{14} Hz$ (b) $K_{max} = e V_{o}$ $\frac{h c}{\lambda} - W_{o}= e V_{o}$ $\lambda = \frac{h c}{e V_{o} + W_{o}}$ $\lambda = \frac{6.63 \times 10^{-34}Js \times 3 \times 10^{8} ms^{-}}{0.60 eV + 2.14 eV}$ $\lambda = \frac{19.89 \times 10^{-26} Jm}{2.74 eV} = \frac{19.89 \times 10^{-26} Jm}{2.74 \times 1.6 \times 10^{-19} J}$ $\lambda = 453.7 \times 10^{-9} m = 453.7 nm$ --- ### DUAL NATURE OF MATTER: DE-BROGLIE WAVES The waves associated with material particles in motion are called matter or de Broglie waves and their wavelength is called de Broglie wavelength. de-Broglie's wave equation: Considering photon as a particle of mass $m$, the energy associated with it is given by Einstein's mass-energy relationship as $E=m c^{2}$ $hv = mc^2$ $\frac{hc}{\lambda} = m c^{2}$ $\lambda = \frac{h}{m c} = \frac{h}{p}$ --- K.E. gained by the electron, $K=\frac{1}{2} m v^{2} = \frac{p^2}{2 m} \Rightarrow P=\sqrt{2 m K}$ Work done on the electron $=e v$ Hence the de Broglie wavelength of the electron is $\lambda = \frac{h}{p} =\frac{h}{\sqrt{2 m k}}=\frac{h}{\sqrt{2 m e V}}= \frac{3}{2} K T$ --- ### DE-BROGLIE WAVELENGTH OF AN ELECTRON Now $h = 6.63 \times 10^{-34} Js$ $m = 9.1 \times 10^{-31} kg$ $e = 1.6 \times 10^{-19} C$ $\therefore \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m K}}=\frac{h}{\sqrt{2 m e V}}$ $\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} V}}$ $ \lambda = \frac{12.3 \times 10^{-10}}{\sqrt{V}} m$ $ \lambda = \frac{12.3}{\sqrt{V}} \overset{o}{A} = 150 \sqrt{V} \overset{o}{A}$ --- ### FAILURE OF THOMSON'S MODEL OF AN ATOM Thomson model remained popular till about 1911 and was discarded later on due to the following drawbacks : * It could not explain the origin of several spectral series in the case of hydrogen and other atoms. * It failed to explain the large angle scattering of alpha- particles in Rutherford's experiment. *Diagram of the Thomson Model is shown. It is a positively charged sphere with electrons scattered throughout.* --- ### RUTHERFORD'S ALPHA SCATTERING EXPERIMENT $N_{\theta} \propto \frac{1}{Sin^{4} \frac{\theta}{2}}$ Observations: * Most of the $\alpha$-particles were found to pass through the gold foil without any appreciable deflection. * In passing through the gold foil, the different $\alpha $-particles underwent different amounts of deflections. A large number of $\alpha$-particles suffered fairly large deflections. * A very small number of $\alpha$-particles (about 1 in 8,000) practically retraced their paths or suffered deflection of nearly 180°. * The graph between the total number of a-particles N(0) scattered through angle O and the scattering angle $\theta$ was found to be as shown. *Graphic: A curve starts high up on the y axis and decrease sharply as it moves toward the right.* --- ### DISTANCE OF CLOSEST APPROACH $K_{\alpha} = \frac{1}{4 \pi \epsilon_{o}} \frac{(z e)(2 e)}{r_{o}}$ $r_{o}= \frac{1}{4 \pi \epsilon_{o}} \frac{2 z e^{2}} {K_{\alpha}}$ $K_{\alpha}= 5 \, MeV$ $=5 \times 10^{6} \times 1.6 \times 10^{-19} \, J$ --- ### IMPACT PARAMETER Impact parameter of the alpha particle is defined as the perpendicular distance of the velocity vector of the alpha particle from the centre of the nucleus, when it is far away from the atom. b = $\\frac{1}{4 \\pi \\epsilon_{o}}$ $\\frac{ Z e^{2}cot \\theta2} { \\frac{1}{2} mu^{2}}$ $\\theta=180^{o}$ $=> b =$$\\frac{1}{4 \\pi \\epsilon_{o}}$ $\\frac{ Z e^{2} Cot 90} { K_{\\alpha}}$ b = 0 *Illustrative graphic shows projectile motion by an alpha particle close to the nucleus of an atom.* --- ### DRAWBACKS OF RUTHERFORD'S ATOM MODEL According to electromagnetic theory, an accelerated charged particle must radiate electromagnetic energy. An electron revolving around the nucleus is under continuous acceleration towards the centre. It should continuously lose energy and move in orbits of gradually decreasing radii. The electron should follow a spiral path and finally it should collapse into the nucleus. Thus the Rutherford's model cannot explain the stability of an atom. In Rutherford's model, an electron can revolve in orbits of all possible radii. So it should emit a continuous spectrum. But an atom like hydrogen always emits a discrete line spectrum. --- ### QUESTION Calculate the impact parameter of a 5MeV particle scattered by 90° when it approaches a gold nucleus. Solution. Here $ K = 5MeV - = 5 \ x \ 1.6 \ 10^{-13 J}$ \\theta = 90°, Z = 79 Impact parameter, $\ b = k Z e^{2} \\frac{cot \\frac{ \\theta } {2} } k$ b = $\\frac{ 9 x 10^{9} x 79 x ( 1.6 x 10^{-19})^{2} cot 45^{} } {5 x 1. 6 x 10^{-13} }$ b = 2.27 x $10^{-14 m}$. --- ### BOHR'S QUANTISATION CONDITION According to de Broglie hypothesis, this electron is also associated with wave character. Hence a circular orbit can be taken to be a stationary energy state only if it contains an integral number of de-Broglie wavelengths, $2 \pi r = n \\lambda$ But de Broglie wavelength, $\\lambda = \\frac{h} {mv}$ $\\therefore 2\\pi r = \\frac{nh} {mv}$ The angular momentum L of the electron must be $L = mvr = \\frac{nh} {2\\pi}$ n = 1, 2, 3 .... This is the famous Bohr's quantisation condition for angular momentum. * Thus only those circular orbits can be the allowed stationary states of an electron in which its angular momentum is an integral multiple of $h/2\pi$ --- ### BOHR'S THEORY OF HYDROGEN ATOM $mv^{2} = \\frac{k(ze)(e) } r$(1) $mv^{2} = \\frac{Kze^{2}}{r} $ $r = \\frac{Kze^{2}}{mv^{2}} $(2) $mvr = \\frac{nh}{2\\pi}$ $r = \\frac{n}{2 \\pi mv}$(3) $\\frac{Kze^{2}}{mv^{2} }$ $kze^{2} = \\frac{nh}{2\\pi mv} $ $v = \\frac{2\\pi Kze^{2}}{nh} $ (4) r = $ \\frac{nh}{2\\pi m (\\frac{2\\pi Kze^{2}} {nn}) }$ $r = \\frac{1}{4\\pi^{2}} $ $\\frac{n^{2} h^{2}} {m Kze^{2}}$ $v = \\frac{2 \\pi Kze^{2}}{H} $\\frac{Z} H$ kE=$\\frac{1}{2} mu^{2}$ =$\\frac{Kze^{2}}{2r}$ $PE = \\frac{k(ze)(-e)}{r} $=$\\frac{-Kze^{2}}{r}$ $\r = \\frac{h^{2}}{4\\pi^{2} mKe^{2} Z}$ $\newline \ $$ $TE = \\frac{kze^{2}} {2r}+ $\\frac{-k(ze)^{2}} {2r} = \\frac{kze^{2}}{2r} = -$\\frac{k^{z}e^{2}}{2 (4 \\pi^{2} mke) }$= -2\\pi^{2}m k e^{-8} Z+$ TE=-13.6ev() --- ### BOHR'S THEORY OF HYDROGEN ATOM $r=\\frac{n^{2} h^{2}}{4\\pi^{2} m k Z e^{2}}$ The radius of the innermost orbit of the hydrogen atom, called Bohr's radius can be determined by putting Z = 1 and n = 1 in equation and it is denoted by $r_{o}$ $\\therefore r_{o}= \\frac{h^{2}}{4\\pi^{2} mke^{2}} =\\frac{(6.63\\times 10^{-34})}{72 \\times 9.11 \\times 10^{-31} \\times 9 x 10^{9} \\times (1.6 \\times 10^{-19})^{2}}$ $m = 5.29 \\times 10^{-11} m$ $r_{o} = 0.53\\AA$ --- ### BOHR'S THEORY OF HYDROGEN ATOM For hydrogen, Z = 1, therefore, $\\nu = \\frac{2\\pi kZe^{2}}{nh}$ $\\nu =\\frac{2\\pi ke^{2}}{nh} = (\\frac{2\\pi ke^{2}}{ch}) \\frac{c}{h}$ $\\nu = \\alpha \\cdot \\frac{c}{n}$ The quantity $\\alpha=\\frac{2\\pi ke^{2}} {cn}$ is a dimensionless constant and is called fine structure constant. Its value is $\\alpha = \\frac{2 \\pi \\times 9 \\times 10^{9} \\times (1.6 \\times 10^{-19})}{3 \\times 10^{8} \\times 6.63\\times 10^{-34}}$ $ \\alpha = \\frac{1}{1.37}$ For first orbit (n = 1), $v= \\frac{1} {137} \\frac{c}{137 n }$ $v= \\frac{1} {137} .$ --- ### BOHR'S THEORY OF HYDROGEN ATOM Kinetic energy of the electron in nth orbit is $K.E =1/2 m v^{2}=kZe/(2r)$ Potential energy of the electron in nth orbit is $P.E = k(q_1q_2)r ^{=kZe (-e)_{r =-kZe/_{r}}kZ e^{4\\pi^{2}mkZ e^{2/}n^{2}n^{2}h^{2}}}$ Hence total energy of the electron in nth orbit is $kZe^{2/-/2r /- =Er/n +p.E kZe\\to^{24\\pi2nkZe2/n^4 h*2}E_n-=-2\\pi^{2}mk Z^{2e^{4\\pi^{2}mkZ e^{4-}n^{2}h^{2}}}E_n-}$ **The negative value of the total energy indicates that the electron is bound to the nucleus by means of electrostatic attraction and some work is required to be done to pull it away from the nucleus.** --- *Illustrative graphic of energy transitions of $E_{4}$* --- ### SPECTRAL SERIES OF HYDROGEN ATOM **(i) Lyman series(UV Region)** If an electron jumps from any higher energy level $n_2=2,3,4...$ to a lower energy level $n_1 =1$, we get a set of spectral lines called Lyman series which belong to the ultraviolet region of the electromagnetic spectrum. This series is given by $\\bar{\\nu=}$=R [$(1/12) -1/[n^{2}]], n_2=2 ,3,4,.$ Rydberg's Constant R = 1.09\\times [10{7 m^1}]$\\to/10^ $\\frac{1}{\\lambda max}$=[R ($\\frac{1}{1^2) -1/(n^{2}]]$