A) 97 nm
B) 82 nm
C) 67 nm
D) 107 nm
Correct Answer: A
Solution :
As, \[E=\frac{-13.6}{{{n}^{2}}}eV\] For \[n=1,\,\,{{E}_{1}}=\frac{-13.6}{{{1}^{2}}}eV=-13.6\,eV\] For \[n=4,\,\,{{E}_{2}}=\frac{-13.6}{{{4}^{2}}}eV=-0.85\,eV\] Now, \[\Delta E={{E}_{2}}-{{E}_{1}}=-0.85-(-13.6)\] \[=12.75\,eV\] \[=2.04\times {{10}^{-18}}J\] This energy is equivalent to energy of photon i.e. \[2.04\times {{10}^{-18}}=\frac{hc}{\lambda }\] \[\Rightarrow \] \[\lambda =\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2.04\times {{10}^{-18}}}=97\,nm\]
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