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 \[\upsilon =\frac{2\pi KZ{{e}^{2}}}{nh}\] \[\therefore \,\,\frac{{{\upsilon }_{1}}}{{{\upsilon }_{2}}}=\frac{{{n}_{2}}}{{{n}_{1}}}=\frac{4}{3}\]You need to login to perform this action.
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