","comment":{"@type":"Comment","text":" $n = 1$ represents the Lyman series, $n = 2$ represents the Balmer series, $n = 3$ represents the Paschen series, $n = 4$ represents Brackett series and $n = 5$ represents the Pfund series."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$A$$ – Pfund series, $$B$$ – Lyman series, $$C$$ – Paschen series, $$D$$ – Balmer series. ","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$A$$ – Brackett series, $$B$$ – Lyman series, $$C$$ – Pfund series, $$D$$ – Balmer series.","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$A$$ – Lyman series, $$B$$ – Balmer series, $$C$$ – Paschen series, $$D$$ – Pfund series. ","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$A$$ – Lyman series, $$B$$ – Balmer series, $$C$$ – Paschen series, $$D$$ – Brackett series. ","position":0,"answerExplanation":{"@type":"Comment","text":" When an electron jumps from all the energy levels to the $$n = 1$$ energy level, the lines fall into the ultra-violet region and those lines are called a Lyman series. Similarly, the lines of the Balmer series are obtained when the electron jumps to $$n = 2$$ energy level and so on.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Bohr model Quiz 1","text":" According to the generalised formula stated by Bohr’s model, we know the energy of an electron in an $$H$$ atom in the ground state is $$13.6 eV$$. What will be its energy in the second excited state? ","comment":{"@type":"Comment","text":" This problem can be easily solved by the equation of energy of an electron given by Bohr’s atomic model i.e. ${{E}_{n}}=-\\dfrac{{{E}_{1}}}{{{n}^{2}}}eV$."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" – 5.6 eV","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" – 2.35 eV","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" – 3.4 eV","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" – 1.51 eV","position":1,"answerExplanation":{"@type":"Comment","text":" $$ \\text{We know that,}\\\\ \\text{Energy of an electron in }{{n}^{th}} \\text{level of H atom is represented as;}\\\\ {{E}_{n}}=-\\dfrac{{{E}_{1}}}{{{n}^{2}}}eV \\text{and we are given that }{{E}_{1}}=13.6eV \\text{thus,}\\\\ \\text{Energy of an electron in second excited state i.e. } n = 3 \\text{ will be,}\\\\ {{E}_{3}}=-\\dfrac{13.6}{{{3}^{2}}}=-\\dfrac{13.6}{9}=-1.51eV\\\\ \\text{Thus, 1.5 eV is the energy of an} \\text{ electron in the second excited state.}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Bohr model Quiz 1","text":" Pritesh is well known for the equation of radius of an orbit in the H – atom by Bohr’s atomic model. Can you help him to find the ratio of the second orbit to that of the third orbit?","comment":{"@type":"Comment","text":" This can be easily solved by the formula for the radius of orbit as described by the Bohr’s atomic model i.e. ${{r}_{n}}=\\dfrac{{{n}^{2}}{{h}^{2}}}{4{{\\Pi }^{2}}m{{e}^{2}}}$."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"$$\\dfrac{9}{4}$$ ","position":1},{"@type":"Answer","encodingFormat":"text/html","text":"$$\\dfrac{6}{2}$$ ","position":2},{"@type":"Answer","encodingFormat":"text/html","text":"$$\\dfrac{2}{6}$$ ","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"$$\\dfrac{4}{9}$$ ","position":0,"answerExplanation":{"@type":"Comment","text":"$$ \\text{We know that the radius of orbit is given by,}\\\\ {{r}_{n}}=\\dfrac{{{n}^{2}}{{h}^{2}}}{4{{\\Pi }^{2}}m{{e}^{2}}} \\\\ {{r}_{n}}\\propto {{n}^{2}} \\\\ \\text{Thus, radius of second orbit is given as } {{r}_{2}}\\propto {{2}^{2}} \\\\ \\text{and radius of third orbit is given as } {{r}_{3}}\\propto {{3}^{2}} .\\\\ \\text{The ratio is given as;}\\\\ \\dfrac{{{r}_{2}}}{{{r}_{3}}}=\\dfrac{4}{9}$$ ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Bohr model Quiz 1","text":" Which of the following has Bohr’s model helped us to calculate with respect to H – atom? $$ \\\\ $$ $$(I)$$ Energy of the electron in particular orbit. $$ \\\\ $$ $$(II)$$ Radius of each circular orbit. $$ \\\\ $$ $$(III)$$ Velocity of electrons in any orbit. $$ \\\\ $$ $$(IV)$$ Number of revolutions made by an electron around the nucleus. $$ \\\\ $$ $$(V)$$ Ionisation energies of H -atom and H like atoms. $$ \\\\ $$ $$(VI)$$ Wavelength of spectral lines. ","comment":{"@type":"Comment","text":" The wavelength of spectral lines is stated by Rydberg and his equation. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" All are correct.","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" None of them are correct.","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" Only $$(VI)$$ is correct.","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" All except $$(VI)$$ are correct.","position":3,"answerExplanation":{"@type":"Comment","text":"Bohr’s atomic model and the spectral analysis of the H – atom has helped us calculate everything described above except the wavelength of spectral lines.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}