","comment":{"@type":"Comment","text":" The intensity of a light beam is the power of the source divided by the area of the cross section on which the light is incident. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$4:\\sqrt{3}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$3:4$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$4:3$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{3}:4$$","position":0,"answerExplanation":{"@type":"Comment","text":" The intensity of the light beam at the cross section is given by; $$ \\\\ I=\\dfrac{P}{A} \\\\ $$ Cross-section area for the first mirror will be the sine of the area of the mirror. So, cross-section for the first mirror is given by; $$ \\\\ \\Rightarrow {{A}_{1}}={{a}^{2}}\\sin {{60}^{\\circ }} \\\\ \\Rightarrow {{A}_{1}}=\\dfrac{{{a}^{2}}\\sqrt{3}}{2} \\\\ $$ Intensity at the first mirror is given by; $$ \\\\ \\Rightarrow {{I}_{1}}=\\dfrac{P}{{{A}_{1}}}=\\dfrac{2P}{{{a}^{2}}\\sqrt{3}}\\text{ }\\to \\left( 1 \\right) \\\\ $$ Cross-section area of the second mirror is given as the cosine of the effective horizontal area of the first mirror; $$ \\\\ $$ So, cross-section of the second mirror will be given as; $$ \\\\ \\Rightarrow {{A}_{2}}={{A}_{1}}\\cos {{60}^{\\circ }}\\cos {{30}^{\\circ }} \\\\ \\Rightarrow {{A}_{2}}=\\dfrac{{{a}^{2}}\\sqrt{3}}{2}\\left( \\dfrac{1}{2} \\right)\\left( \\dfrac{\\sqrt{3}}{2} \\right)=\\dfrac{3{{a}^{2}}}{8} \\\\ $$ The intensity at the second mirror is given by; $$ \\\\ \\Rightarrow {{I}_{2}}=\\dfrac{P}{{{A}_{2}}} \\\\ \\Rightarrow {{I}_{2}}=\\dfrac{8P}{3{{a}^{2}}}\\text{ }\\to \\left( 2 \\right) \\\\ \\text{Dividing the equations 1 and 2} \\\\ \\Rightarrow \\dfrac{{{I}_{1}}}{{{I}_{2}}}=\\dfrac{\\left( \\dfrac{2P}{{{a}^{2}}\\sqrt{3}} \\right)}{\\left( \\dfrac{8P}{3{{a}^{2}}} \\right)}=\\dfrac{2\\sqrt{3}}{8} \\\\ \\Rightarrow \\dfrac{{{I}_{1}}}{{{I}_{2}}}=\\dfrac{\\sqrt{3}}{4} \\\\ \\therefore {{I}_{1}}:{{I}_{2}}=\\sqrt{3}:4$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Quantum mechanics Quiz 1","text":" A hydrogen like atom can absorb an energy of $$690\\text{ eV}\\\\$$. The emission spectrum is shown in the diagram below. Find the minimum wavelength of the emitted radiation. $$ \\\\ $$ ","comment":{"@type":"Comment","text":" The difference in the inverse squares of the excitation states is proportional to the inverse of corresponding wavelength. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{9}{100.8{Z}^{2}}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{13.6}{108.8{Z}^{2}}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" None of the above","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{9}{108.8{Z}^{2}}$$","position":1,"answerExplanation":{"@type":"Comment","text":" $$\\text{Wave number is given by;} \\\\ \\bar{\\nu }=\\dfrac{1}{\\lambda }=13.6{{Z}^{2}}\\left( \\dfrac{1}{n_{f}^{2}}-\\dfrac{1}{n_{i}^{2}} \\right) \\\\ \\text{Minimum wavelength is;} \\\\ \\Rightarrow \\dfrac{1}{{{\\lambda }_{\\min }}}=13.6{{Z}^{2}}\\left( \\dfrac{1}{{{1}^{2}}}-\\dfrac{1}{{{3}^{2}}} \\right)=13.6{{Z}^{2}}\\left( \\dfrac{1}{1}-\\dfrac{1}{9} \\right) \\\\ \\Rightarrow \\dfrac{1}{{{\\lambda }_{\\min }}}=13.6{{Z}^{2}}\\left( \\dfrac{9-1}{9} \\right)=\\dfrac{108.8{{Z}^{2}}}{9} \\\\ \\therefore {{\\lambda }_{\\min }}=\\dfrac{9}{108.8{{Z}^{2}}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Quantum mechanics Quiz 1","text":" For a hypothetical hydrogen like atom the difference between the energies in ninth and eighth orbitals is $$8\\text{ eV}\\\\$$. What is the atomic number of this atom?","comment":{"@type":"Comment","text":" The excitation potential is actually the energy difference between the orbital that the electron is exciting to and the initial orbital. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$10$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$12$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$14$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$13$$","position":2,"answerExplanation":{"@type":"Comment","text":" $$ \\text{The energy of the electron in the }n^{th}\\text{ is given by;} \\\\ {{E}_{n}}=\\dfrac{-13.6{{Z}^{2}}}{{{n}^{2}}} \\\\ \\text{Given that;} \\\\ \\Delta {{E}_{\\left( 9-8 \\right)}}=8\\text{ eV} \\\\ \\text{The energy difference between the }9^{th}\\text{ and }8^{th}\\text{ orbitals is given by;} \\\\ \\Delta {{E}_{\\left( 9-8 \\right)}}={{E}_{9}}-{{E}_{8}} \\\\ \\Rightarrow \\Delta {{E}_{\\left( 9-8 \\right)}}=\\left( \\dfrac{-13.6{{Z}^{2}}}{{{9}^{2}}} \\right)-\\left( \\dfrac{-13.6{{Z}^{2}}}{{{8}^{2}}} \\right)=8 \\\\ \\Rightarrow 13.6{{Z}^{2}}\\left( \\dfrac{81-64}{64\\times 81} \\right)=8 \\\\ \\Rightarrow 13.6{{Z}^{2}}=\\dfrac{8\\times 64\\times 81}{17} \\\\ \\Rightarrow Z=\\sqrt{\\dfrac{41472}{231.2}}=\\sqrt{179.4} \\\\ \\therefore Z\\approx 13$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Quantum mechanics Quiz 1","text":" A photocell is made up of a certain material such that when a monochromatic light of wavelength $$6000\\text{ }\\overset{\\circ }{\\mathop{\\text{A}}}\\,$$ is incident on it the maximum velocity reached by the emitted photoelectrons is $$v$$ , but when a monochromatic light of wavelength $$8000\\text{ }\\overset{\\circ }{\\mathop{\\text{A}}}\\,$$ is incident on it the maximum velocity of the emitted photoelectron is $$2v$$ . Find the threshold wavelength of the material. ","comment":{"@type":"Comment","text":" The threshold wavelength is the minimum wavelength for which the emission of photoelectrons will take place. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$5438.49\\text{ }\\overset{\\circ }{\\mathop{\\text{A}}}\\,$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$5439.48\\text{ }\\overset{\\circ }{\\mathop{\\text{A}}}\\,$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$5539.47\\text{ }\\overset{\\circ }{\\mathop{\\text{A}}}\\,$$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$5538.46\\text{ }\\overset{\\circ }{\\mathop{\\text{A}}}\\,$$","position":3,"answerExplanation":{"@type":"Comment","text":" $$\\text{The Einstein }\\!\\!'\\!\\!\\text{ s equation for photoelectron is given by;} \\\\ {{K}_{\\max }}=hf-{{\\omega }_{0}} \\\\ \\text{Given that;} \\\\ {{v}_{1}}=v;\\text{ }{{v}_{2}}=2v\\,\\Rightarrow \\,{{K}_{1}}=K;\\,{{K}_{2}}=4K \\\\ \\text{Substituting the values in the Einstein }\\!\\!'\\!\\!\\text{ s equation;} \\\\ K={{\\left( hf \\right)}_{1}}-h{{f}_{0}} \\\\ \\text{Given that;} \\\\ {{\\left( hf \\right)}_{1}}=\\dfrac{12400}{\\lambda }=\\dfrac{12400}{6000}=\\dfrac{31}{15} \\\\ \\Rightarrow K=\\dfrac{31}{15}-h{{f}_{0}}\\text{ }\\to \\left( 1 \\right) \\\\ \\text{For second light;} \\\\ 2K={{\\left( hf \\right)}_{2}}-h{{f}_{0}} \\\\ {{\\left( hf \\right)}_{2}}=\\dfrac{12400}{\\lambda }=\\dfrac{12400}{8000}=\\dfrac{31}{20} \\\\ \\Rightarrow 4K=\\dfrac{31}{20}-h{{f}_{0}}\\text{ }\\to \\left( 2 \\right) \\\\ \\text{Subtracting equation 1 from equation 2;} \\\\ \\Rightarrow 3K=\\dfrac{31}{20}-\\dfrac{31}{15}=31\\left( \\dfrac{15-20}{300} \\right)=\\dfrac{-31}{60} \\\\ \\Rightarrow K=-\\dfrac{31}{180} \\\\ \\text{Substituting in equation 1;} \\\\ -\\dfrac{31}{180}=\\dfrac{31}{15}-h{{f}_{0}} \\\\ \\Rightarrow h{{f}_{0}}=31\\left( \\dfrac{12+1}{180} \\right)=\\dfrac{403}{180} \\\\ \\Rightarrow \\dfrac{12400}{{{\\lambda }_{0}}}=\\dfrac{403}{180} \\\\ \\Rightarrow {{\\lambda }_{0}}=\\dfrac{12400\\times 180}{403} \\\\ \\therefore {{\\lambda }_{0}}=5538.46\\,\\overset{\\circ }{\\mathop{\\text{A}}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Quantum mechanics Quiz 1","text":" A photocell is operating in saturation mode with a photocurrent of $$8\\text{ }\\mu \\text{A}$$ . If a monochromatic light of wavelength $$4000\\text{ }\\overset{\\circ}{\\mathop{\\text{A}}}\\,$$ and power $$4\\text{ mW}$$ is incident on the photocell, determine the efficiency of the photocell. ","comment":{"@type":"Comment","text":" The efficiency of a machine is the ratio of output achieved by the input supplied. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 8.1%","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" 81%","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" 0.61%","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 0.061%","position":3,"answerExplanation":{"@type":"Comment","text":" $$\\text{The energy of the light is given by;} \\\\ E=\\dfrac{12400}{\\lambda } \\\\ \\Rightarrow E=\\dfrac{12400}{4000}=3.1\\text{ eV} \\\\ \\Rightarrow E=3.1\\times 1.6\\times {{10}^{-19}} \\\\ \\Rightarrow E=4.96\\times {{10}^{-19}} \\\\ \\text{Rate of incidence of photon }=\\dfrac{P}{E}=\\dfrac{4\\times {{10}^{-3}}}{4.96\\times {{10}^{-19}}} \\\\ \\Rightarrow \\text{Rate of incidence of photon }=8.1\\times {{10}^{16}}\\text{ }{{\\text{s}}^{-1}} \\\\ \\text{Rate of emission of electrons }=\\dfrac{I}{e}=\\dfrac{8\\times {{10}^{-6}}}{1.6\\times {{10}^{-19}}} \\\\ \\Rightarrow \\text{Rate of emission of electrons }=5\\times {{10}^{13}} \\\\ \\text{Efficiency of photocell}\\left( \\gamma \\right)=\\dfrac{\\text{Rate of emission of electrons}}{\\text{Rate of incidence of photons}}\\times 100 \\\\ \\Rightarrow \\gamma =\\dfrac{5\\times {{10}^{13}}}{8.1\\times {{10}^{16}}}\\times 100 \\\\ \\Rightarrow \\gamma =\\dfrac{0.61}{1000}\\times 100 \\\\ \\therefore \\gamma =0.061\\%$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}