A) 4 : 3
B) 3 : 4
C) 2 : 1
D) 1 : 2
Correct Answer: A
Solution :
Number of emitted spectral lines \[N=\frac{n(n-1)}{2}\] 1st case \[N=3\] \[\therefore \] \[3=\frac{{{n}_{1}}({{n}_{1}}-1)}{2}\] \[\Rightarrow \] \[n_{1}^{2}-{{n}_{1}}-6=0\] \[\Rightarrow \] \[({{n}_{1}}-3)({{n}_{1}}+2)=0\] \[{{n}_{1}}=3,{{n}_{1}}=-2\] Negative value of \[{{n}_{1}}\]is not possible \[\therefore \] \[{{n}_{1}}=3\] 2nd case \[N=6\] Again, \[6=\frac{{{n}_{2}}({{n}_{2}}-1)}{2}\] \[\Rightarrow \] \[n_{2}^{2}-{{n}_{2}}-12=0\] \[\Rightarrow \] \[({{n}_{2}}-4)({{n}_{2}}+3)=0\] \[{{n}_{2}}=4,{{n}_{2}}=-3\] Again, as \[{{n}_{2}}\]is always positive \[\therefore \] \[{{n}_{2}}=4\] Velocity of electron \[v=\frac{Z{{e}^{2}}}{2{{\varepsilon }_{0}}hn}\] \[\therefore \] \[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{{{n}_{2}}}{{{n}_{1}}}\] \[\Rightarrow \] \[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{4}{3}\]
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