A) \[{{\lambda }_{3}}={{\lambda }_{1}}+{{\lambda }_{2}}\]
B) \[{{\lambda }_{3}}=\frac{{{\lambda }_{1}}{{\lambda }_{2}}}{{{\lambda }_{1}}+{{\lambda }_{2}}}\]
C) \[{{\lambda }_{1}}+{{\lambda }_{2}}+{{\lambda }_{3}}=0\]
D) \[\lambda _{3}^{2}=\lambda _{1}^{2}+\lambda _{2}^{2}\]
Correct Answer: B
Solution :
Let the energy in A, B and C state be EA. EB and EC, then from the figure \[{{\nu }_{\text{Balmer}}}=\frac{c}{{{\lambda }_{\max }}}=Rc\left[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(3)}^{2}}} \right]=\frac{5RC}{36}\]or \[\frac{hc}{{{\lambda }_{1}}}+\frac{hc}{{{\lambda }_{2}}}=\frac{hc}{{{\lambda }_{3}}}\] \[\Rightarrow \,{{\lambda }_{3}}=\frac{{{\lambda }_{1}}{{\lambda }_{2}}}{{{\lambda }_{1}}+{{\lambda }_{2}}}\]You need to login to perform this action.
You will be redirected in
3 sec