A) \[{{\text{n}}_{\text{1}}}=\text{ 6 to }{{\text{n}}_{2}}=\text{2}\]
B) \[{{\text{n}}_{1}}=\text{6 to }{{\text{n}}_{2}}=1\]
C) \[{{\text{n}}_{\text{1}}}=\text{2 to }{{\text{n}}_{2}}=\text{6}\]
D) \[{{\text{n}}_{\text{1}}}=1\text{ to }{{\text{n}}_{2}}=2\]
Correct Answer: A
Solution :
From Bohrs concept, the wavelength \[(\lambda )\] of emitted light is given by \[\frac{1}{\lambda }=R\left( \frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}} \right)\] For \[{{n}_{1}}=1,\,\,\,{{n}_{2}}=2\] \[\therefore \] \[\frac{1}{{{\lambda }_{1}}}=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)\] \[=R\left( \frac{1}{4}-\frac{1}{4} \right)=0.75\,R\] For \[{{n}_{1}}=2,\,\,\,{{n}_{2}}=6\] \[\therefore \] \[\frac{1}{{{\lambda }_{2}}}=R\left( \frac{1}{4}-\frac{1}{36} \right)=0.22R\] Also frequency \[(v)\propto \frac{1}{\lambda }\]. and E = h v Hence, for minimum E, \[{{n}_{2}}=6\] and \[{{n}_{1}}=2\].
You need to login to perform this action.
You will be redirected in
3 sec
