A) \[300A{}^\circ \]
B) \[2.5\times {{10}^{-11}}m\]
C) \[100A{}^\circ \]
D) \[1.5\times {{10}^{-11}}m\]
Correct Answer: A
Solution :
[a] \[{{E}_{n}}=-\frac{I.E.}{{{n}^{2}}}\] for Bohr "s hydrogen atom. Here, \[I.E/=4R\text{ }\therefore {{E}_{n}}=\frac{-4R}{{{n}^{2}}}\] \[\therefore {{E}_{2}}-{{E}_{1}}=\frac{-4R}{{{2}^{2}}}-\left( -\frac{4R}{{{l}^{2}}} \right)=3R\] ?.(i) \[{{E}_{2}}-{{E}_{1}}=hv=\frac{hc}{\lambda }\] ?(ii) From (i) and (ii) \[\therefore \lambda =\frac{hc}{3R}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2.2\times {{10}^{-18}}\times 3}=300\overset{\text{o}}{\mathop{\text{A}}}\,\]
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