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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 40

  • question_answer
    A He-atom is de-excited from an energy level "n" to ground state to emit two consecutive photons of wavelength \[1085\overset{\text{o}}{\mathop{\text{A}}}\,\]  and\[3040\overset{o}{\mathop{\text{A}}}\,\]. Then n will be-

    A) 3                                 

    B) 4    

    C) 5         

    D) 6

    Correct Answer: C

    Solution :

    [c] \[E=-13.6\,{{Z}^{2}}/{{n}^{2}}=-\frac{13.6\times 4}{{{n}^{2}}}=\frac{54.4}{{{n}^{2}}}eV\] In coming to ground state \[\Delta E={{E}_{0}}-{{E}_{i}}\] \[=\left( \frac{-54.4}{{{n}^{2}}} \right)-\left( \frac{-54.4}{1} \right)\] \[=54.4\left( 1-\frac{1}{{{n}^{2}}} \right)eV\]                            ?. (1) But \[\Delta E={{E}_{ph}}=\frac{hc}{{{\lambda }_{1}}}+\frac{hc}{{{\lambda }_{2}}}\] \[=12400\left( \frac{1}{1085}+\frac{1}{304} \right)eV\] ?. (2) From (1) & (2) \[=54.4\left( 1-\frac{1}{{{n}^{2}}} \right)=52.08\] \[\Rightarrow 1-\frac{-1}{{{n}^{2}}}=0.96\Rightarrow n=5\]                  


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