A) \[1.54\times l{{0}^{15}}\text{ }{{s}^{-1}}\]
B) \[1.03\times {{10}^{15}}\,\,{{s}^{-1}}\]
C) \[3.08\times {{10}^{15}}\text{ }{{s}^{-1}}\]
D) \[2.00\times {{10}^{15}}\text{ }{{s}^{-1}}\]
Correct Answer: C
Solution :
| Ionization energy of \[H=2.18\times {{10}^{-18}}J\,ato{{m}^{-1}}\] |
| \[\therefore \] \[{{E}_{1}}\] (Energy of 1st orbit of H-atom) |
| \[=-2.18\times l{{0}^{-18}}\text{ }J-ato{{n}^{-1}}\] |
| \[\therefore \] \[{{E}_{n}}=\frac{-2.18\times {{10}^{18}}}{{{n}^{2}}}J-ato{{m}^{-1}}\] |
| \[Z=1\] for \[H-atom\] |
| \[\Delta E={{E}_{4}}-{{E}_{1}}\] |
| \[=\frac{-2.18\times {{10}^{-18}}}{{{4}^{2}}}-\frac{-2.18\times {{10}^{-18}}}{{}}\] |
| \[=-2.18\times {{10}^{-18}}\times \left[ \frac{1}{{{4}^{2}}}-\frac{1}{{{1}^{2}}} \right]\] |
| \[\Delta E=hv=-2.18\times {{10}^{-18}}\times -\frac{15}{16}\] |
| \[=+\,2.0437\times {{10}^{-18}}J\,ato{{m}^{-1}}\] |
| \[\therefore \] \[V=\frac{\Delta E}{h}=\frac{2.0437\times {{10}^{-18}}J\,ato{{m}^{-1}}}{6.625\times {{10}^{-34}}\,J\,\,s}\] |
| \[=3.084\times {{10}^{15}}\text{ }{{s}^{-1}}\text{ }ato{{m}^{-1}}\] |
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