A) \[1.54\times {{10}^{15}}{{s}^{-1}}\]
B) \[1.03\times {{10}^{15}}{{s}^{-1}}\]
C) \[3.08\times {{10}^{15}}{{s}^{-1}}\]
D) \[2.00\times {{10}^{15}}{{s}^{-1}}\]
Correct Answer: C
Solution :
lonisation energy of H \[=2.18\times {{10}^{-18}}J\text{ }ato{{m}^{-1}}\] \[\therefore \]\[{{E}_{1}}\] (Energy of 1st orbit of H-atom) \[=-2.18\times {{10}^{-18}}J-ato{{m}^{-1}}\] \[\therefore \] \[{{E}_{n}}=\frac{-2.18\times {{10}^{-18}}}{{{n}^{2}}}J-ato{{m}^{-1}}\] \[Z=1\]for H-atom \[\Delta E={{E}_{4}}-{{E}_{1}}\] \[=\frac{-2.18\times {{10}^{-18}}}{{{4}^{2}}}-\frac{-2.18\times {{10}^{-18}}}{{{1}^{2}}}\] \[=-2.18\times {{10}^{-18}}\times \left[ \frac{1}{{{4}^{2}}}-\frac{1}{{{1}^{2}}} \right]\] \[\Delta E=hv=-2.18\times {{10}^{-18}}\times -\frac{15}{16}\] \[=+2.0437\times {{10}^{-18}}J\,ato{{m}^{-1}}\] \[\therefore \]\[v=\frac{\Delta E}{h}=\frac{2.0437\times {{10}^{-18}}J\,ato{{m}^{-1}}}{6.625\times {{10}^{-34}}Js}\] \[=3.084\times {{10}^{15}}{{s}^{-1}}ato{{m}^{-1}}\]
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