$$\\\\ \\text{The path difference between electrons scattered from adjacent crystal planes is}\\\\ dsin\\theta +dsin\\theta = 2dsin\\theta\\\\ \\text{We are given that}\\,d =0.091, \\theta = 65^{\\circ}\\,\\text{and for intensity maxima, the point of constructive interference n=1.}\\\\ \\text{Now, for constructive interference to happen between the two scattered beams to produce an intensity maxima,}\\\\ \\text{the path difference must be an integer multiple of the electron wavelength, i.e.,}\\\\ 2dsin\\theta = n\\lambda\\\\ \\text{This is nothing but the Bragg’s Wave Equation.}\\\\ 2\\times 0.091\\times 10^{-9} \\times sin(65^{\\circ}) = 1\\times \\lambda\\\\ \\Rightarrow \\lambda = 0.1656 \\times 10^{-9}\\,m\\\\ \\therefore\\lambda = 0.1656\\,nm$$ ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Modern physics Quiz 1","text":" Fill in the blank: Robert Millikan, much like the other physicists in 1916, was never an advocate of the particle theory of light since the wave nature of light had already been proved by the Michelson Morley experiment, and looked at Einstein’s photoelectric equation with much scrutiny. To this end, he decided to arrive at conclusive results by performing a vacuum sealed photoelectric effect experiment himself. Being an experimental physicist, his procedure was thorough and his findings ended up corroborating Einstein’s theorized photoelectric equation instead of disproving its validity since he was able to verify the value of Planck’s constant to be nearly ___ $$erg\\,s$$ which was acceptable at the time. He did so by plotting a volt-frequency graph similar to the following figure. $$(\\text{Take photoelectron charge to be e =} 1.59 \\times 10^{-19}\\,J)$$ ","comment":{"@type":"Comment","text":"Find the slope of the photoelectric equation and equate that to the slope of the graph. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$6.86 \\times 10^{-34}\\,Js$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$6.62 \\times 10^{-27}\\,Js$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$6.62 \\times 10^{-34}\\,erg\\,s$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$6.56 \\times 10^{-27}\\,erg\\,s$$","position":1,"answerExplanation":{"@type":"Comment","text":" $$ \\text{According to Einstein’s photoelectric equation:}\\\\ K_{max} = h\\nu - \\phi_0,\\\\ \\text{But the maximum kinetic energy of a potentially accelerated particle can be given as:}\\\\ K_{max} = eV,\\\\ \\text{From the above two equations, we have:}\\\\ eV = h\\nu - \\phi_0\\\\ \\Rightarrow V = \\left(\\dfrac{h}{e}\\right)\\nu-\\dfrac{\\phi}{e}\\\\ \\text{The slope of the above equation is:}\\\\ m = \\dfrac{h}{e}\\\\ \\text{But from the graph, the slope of the straight line can be calculated as:}\\\\ m = \\dfrac{V_2-V_2}{\\nu_2-\\nu_1}\\\\ \\Rightarrow m = \\dfrac{1-(-2)}{(121-48.25) \\times 10^{13}}\\\\ \\Rightarrow m = 4.124 \\times 10^{-15}\\\\ \\text{Equating the two slope expressions, we get:}\\\\ \\dfrac{h}{e} = 4.124 \\times 10^{-15}\\\\ \\Rightarrow h = 1.59 \\times 10^{-19} \\times 4.124 \\times 10^{-15}\\\\ \\Rightarrow h = 6.557 \\times 10^{-34}\\,J\\,s\\\\ \\therefore h = 6.56 \\times 10^{-27}\\,erg\\,s$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Modern physics Quiz 1","text":" Estimate the number of mean lives elapsed when the number of atoms in a radioactive sample reduces to 20% the original value. $$(\\text{Take}\\, ln\\,5 = 1.609)$$ ","comment":{"@type":"Comment","text":" Find the elapsed time in terms of the decay constant and number of mean lives and plug it into the radioactive decay equation. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 3","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" 1","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" 4","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 2 ","position":0,"answerExplanation":{"@type":"Comment","text":"$$ \\text{Let}\\,N_0\\,\\text{be the initial amount of radioactive sample.}\\\\ \\text{We are given that}\\,N = 20\\% \\,of \\,N_0 = \\dfrac{20}{100}N_0 = \\dfrac{N_0}{5}\\\\ \\text{But from the equation of radioactive decay:}\\\\ N = N_0 e^{-\\lambda t}\\\\ \\Rightarrow \\dfrac{N_0}{5} = N_0 e^{-\\lambda t}\\\\ \\text{Now, let the time at which the sample reduces to}\\,20\\%\\,\\text{its initial value be t and let the number of mean lives elapsed be m, i.e.,}\\\\ t = m\\tau,\\\\ \\text{We know that mean life}\\,\\tau = \\dfrac{1}{\\lambda}\\\\ \\Rightarrow t = \\dfrac{m}{\\lambda}\\\\ \\text{Plugging this into the radioactive decay equation, we get:}\\\\ \\dfrac{N_0}{5} = N_0 e^{-\\lambda. \\dfrac{m}{\\lambda}}\\\\ \\Rightarrow e^{-m} = \\dfrac{1}{5}\\\\ \\Rightarrow e^m = 5\\\\ \\text{Taking logarithm on both sides:}\\\\ \\therefore m = ln\\,5 = 1.609 \\approx 2$$ ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Modern physics Quiz 1","text":" Whenever a string of length $$L$$ is vibrated, it emits harmonics of only definite wavelengths $$n\\lambda$$. Similarly, a free particle of mass $$m$$ moving with a velocity $$v$$ is associated with its de Broglie wavelength $$\\lambda$$ by virtue of its motion. What would be the kinetic energy of this free particle when it is confined to move in a straight line of length $$L$$? ","comment":{"@type":"Comment","text":"Equate the wavelength equations for harmonics and de Broglie wavelength to arrive at a parameter that you can substitute in the classical kinetic energy equation and solve. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"$$\\dfrac{mv^2}{2m}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":"$$\\dfrac{8mL^2}{n^2h^2}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":"$$\\dfrac{nh}{2mL}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"$$\\dfrac{n^2h^2}{8mL^2}$$","position":2,"answerExplanation":{"@type":"Comment","text":"$$ \\text{For a vibrating string of length L, the wavelengths at which it emits harmonics is given by:}\\\\ n\\lambda = 2L\\\\ \\Rightarrow \\lambda = \\dfrac{2L}{n},\\\\ \\text{For a free particle of mass m moving with a velocity v, the de Broglie wavelength associated with it is given by:}\\\\ \\lambda = \\dfrac{h}{p} = \\dfrac{h}{mv}\\\\ \\text{Now, just like the string, when this particle is confined to move on a line of length L,}\\\\ \\text{it can only have certain discrete values of}\\,\\lambda\\,\\text{given by the vibrating string equation we have above, i.e.,}\\\\ \\lambda = \\dfrac{2L}{n}\\\\ \\Rightarrow \\dfrac{h}{mv} = \\dfrac{2L}{n}\\\\ \\Rightarrow mv = \\dfrac{nh}{2L}\\\\ \\text{Now, the kinetic energy of this free particle is given as:}\\\\ KE = \\dfrac{1}{2}mv^2\\\\ \\text{Multiplying the numerator and denominator of RHS by m:}\\\\ KE = \\dfrac{(mv)^2}{2m}\\\\ \\Rightarrow KE = \\dfrac{\\left(\\dfrac{nh}{2L}\\right)^2}{2m}\\\\ \\therefore KE= \\dfrac{n^2h^2}{8mL^2}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Modern physics Quiz 1","text":"In Rutherford’s alpha particle scattering experiment, a nucleus $$N$$ having a charge $$Ze$$ scatters an alpha particle of mass $$m$$ and charges $$2e$$ as shown in the figure. What would be the relation between its impact parameter $$b$$ and the minimum distance $$s$$ from the nucleus? What would be the minimum distance in case of a head-on collision? Take $$KE_{i}$$ as the initial kinetic energy of the alpha particle, and the nuclear electric potential as $$k\\dfrac{Ze}{s}$$.
","comment":{"@type":"Comment","text":"Use the conservation of angular momentum and energy to arrive at an expression for the minimum distance. Remember that for head-on collision the impact parameter is zero. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$s^2 = k\\dfrac{2Ze^2s}{KE_{i}} -b^2$$ and $$s = k\\dfrac{2Ze^2}{KE_{i}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$s^2 = k\\dfrac{2Ze^2s}{KE_{i}}+b$$ and $$s = b$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$s^2 = k\\dfrac{2Ze^2s}{KE_{i}}-b$$ and $$s=b^2$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"$$ s^2 = k\\dfrac{2Ze^2s}{KE_{i}} +b^2$$ and $$s = k\\dfrac{2Ze^2}{KE_{i}}$$","position":0,"answerExplanation":{"@type":"Comment","text":" $$ \\text{The presence of the nucleus gives rise to an electric potential of}\\,V = k\\dfrac{Ze}{s}.\\\\ \\text{This potential accelerates the alpha particle upon incidence with an energy of}\\,U = k\\dfrac{Ze.2e}{s}\\\\ \\text{Applying the law of conservation of angular momentum:}\\\\ (mv)b = (mv^{\\prime})s\\\\ \\Rightarrow v^{\\prime} = \\dfrac{vb}{s}\\\\ \\text{Applying the law of conservation of energy:}\\\\ KE_{i} = KE_{after\\,scattering} + PE_{at\\,s}\\\\ \\dfrac{1}{2}mv^2 = \\dfrac{1}{2}m{v^{\\prime}}^2 + k\\dfrac{Ze.2e}{s}\\\\ \\text{Plugging in the expression for}\\,v^{\\prime}\\,\\text{that we got from momentum conservation:}\\\\ \\dfrac{1}{2}mv^2 = \\dfrac{1}{2}m\\left(\\dfrac{v^2b^2}{s^2}\\right) + k\\dfrac{2Ze^2}{s}\\\\ \\Rightarrow KE_{i}\\left(1-\\dfrac{b^2}{s^2}\\right) = k\\dfrac{2Ze^2}{s}\\\\ \\Rightarrow \\left(\\dfrac{s^2-b^2}{s^2}\\right) k\\dfrac{2Ze^2}{s.KE_{i}}\\\\ \\text{Multiply the whole equation by}\\,s^2:\\\\ s^2-b^2 = k\\dfrac{2Ze^2s}{KE_{i}}\\\\ \\Rightarrow s^2 = k\\dfrac{2Ze^2s}{KE_{i}} +b^2\\\\ \\text{In case of a head-on collision, i.e., when b=0, the above equation becomes}\\\\ s^2 = k\\dfrac{2Ze^2s}{KE_{i})}\\\\ \\Rightarrow s = k\\dfrac{2Ze^2}{KE_{i}}\\\\ \\text{We see that the above expression is equal to the distance of closest approach}\\,r_0\\,\\\\ \\text{which is generally given as:}\\\\ \\therefore r_0 =k\\dfrac{2Ze^2}{KE_{i}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}