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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 37

  • question_answer
    Electrons in a certain energy level \[n\text{ }=\text{ }{{n}_{1}}\], can emit 3 spectral lines. When they are in another energy level, \[n\text{ }=\text{ }{{n}_{2}}\]. They can emit 6 spectral lines. The orbital speed of the electrons in the two orbits are in the ratio of

    A) \[4:3\]   

    B)        \[3:4\]

    C) \[2:1\]               

    D)        \[1:2\]

    Correct Answer: A

    Solution :

    Number of emission spectral lines \[N=\frac{n(n-1)}{2}\] \[\therefore \,\,3=\frac{{{n}_{1}}({{n}_{1}}-1)}{2}\], in first case. Or \[\operatorname{n}_{1}^{2}\,-{{n}_{1}}-6=0\,\,or\,\,({{n}_{1}}-3)({{n}_{1}}+2)=0\] Take positive root. \[\therefore \,\,\,{{n}_{1}}=3\] Again, \[6=\frac{{{n}_{2}}({{n}_{2}}-1)}{2}\], in second case. Or \[\operatorname{n}_{2}^{2}-{{n}_{2}}-12=0\,\,or\,\,({{n}_{2}}\,-4)\,({{n}_{2}}+3)=0\]. Take positive root, or \[{{\operatorname{n}}_{2}}=4\] Now velocity of electron \[\upsilon =\frac{2\pi KZ{{e}^{2}}}{nh}\] \[\therefore \,\,\frac{{{\upsilon }_{1}}}{{{\upsilon }_{2}}}=\frac{{{n}_{2}}}{{{n}_{1}}}=\frac{4}{3}\]


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