A) 110 nm
B) 120 nm
C) 122 nm
D) 125 nm
Correct Answer: C
Solution :
As \[{{E}_{n}}=\frac{-13.6}{{{n}^{3}}}eV\] At ground level \[(n=1),\] \[{{E}_{1}}=\frac{-13.6}{12}=-13.6eV\] At first excited state\[(n=2),\] \[{{E}_{2}}=\frac{-13.6}{{{2}^{2}}}=-3.4eV\] As \[hv={{E}_{2}}-{{E}_{1}}=-34+13.6=10.2eV\] \[=16\times {{10}^{-19}}\times 10.2=1.63\times {{10}^{-18}}J\] Also \[c=v\lambda \] So \[\lambda =\frac{c}{v}=\frac{ch}{{{E}_{2}}-{{E}_{1}}}\] \[=\frac{(3\times {{10}^{8}})\times (6.63\times {{10}^{-34}})}{1.63\times {{10}^{-18}}}\] \[=1.22\times {{10}^{-7}}m\approx 122nm\]
You need to login to perform this action.
You will be redirected in
3 sec
