A) 1.25
B) 0.25
C) 5
D) 10
Correct Answer: B
Solution :
The series end of Lyman series corresponds to transition from \[{{n}_{i}}=\infty \]to \[{{n}_{f}}=1,\]corresponding to the wavelength \[\frac{1}{{{({{\lambda }_{min}})}_{L}}}=R\left[ \frac{1}{1}-\frac{1}{\infty } \right]=R\] \[\Rightarrow \] \[{{({{\lambda }_{min}})}_{L}}=\frac{1}{R}=912{\AA}\] ?(i) For last line of Balmer series \[\frac{1}{{{({{\lambda }_{min}})}_{B}}}=R\left[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(\infty )}^{2}}} \right]=\frac{R}{4}\] \[\Rightarrow \]\[{{({{\lambda }_{min}})}_{B}}=\frac{4}{R}=3636{\AA}\] ?(ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{{{({{\lambda }_{min}})}_{L}}}{{{({{\lambda }_{min}})}_{B}}}=0.25\]
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