A) 5 : 9
B) 5 : 36
C) 1 : 4
D) 3 : 4
Correct Answer: A
Solution :
For maximum wavelength of Balmer series \[\frac{1}{{{\lambda }_{\max }}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{R\times 5}{36}\] ? (i) For minimum wavelength of Balmer series, \[\frac{1}{{{\lambda }_{\min }}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{\infty } \right)=\frac{R}{4}\] ? (ii) From Eqs. (i) and (ii), we have \[\therefore \] \[\frac{{{\lambda }_{\min }}}{{{\lambda }_{\max }}}=\frac{R\times 5}{36}\times \frac{4}{R}=\frac{5}{9}\]
You need to login to perform this action.
You will be redirected in
3 sec
