A) \[\text{1}.\text{54}\times \text{l}{{0}^{\text{15}}}\text{ }{{\text{s}}^{-\text{1}}}\]
B) \[1.03\times {{10}^{15}}\,\,{{s}^{-1}}\]
C) \[\text{3}.0\text{8}\times \text{1}{{0}^{\text{15}}}\text{ }{{\text{s}}^{-\text{1}}}\]
D) \[\text{2}.00\times \text{1}{{0}^{\text{15}}}\text{ }{{\text{s}}^{-\text{1}}}\]
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) \[=-\text{2}.\text{18}\times \text{l}{{0}^{-\text{18}}}\text{ J}-\text{ato}{{\text{n}}^{-1}}\] \[\therefore {{E}_{n}}=\frac{-2.18\times {{10}^{18}}}{{{n}^{2}}}J-ato{{m}^{-1}}\] \[\text{Z}=\text{1 for H}-\text{atom}\] \[\Delta E={{E}_{4}}-{{E}_{1}}\] \[=\frac{-2.18\times {{10}^{-18}}}{{{4}^{2}}}-\frac{-2.18\times {{10}^{-18}}}{{}}\] \[=-2.18\times {{10}^{-18}}\times \left[ \frac{1}{{{4}^{2}}}-\frac{1}{{{1}^{2}}} \right]\] \[\Delta E=hv=-2.18\times {{10}^{-18}}\times -\frac{15}{16}\] \[=+2.0437\times {{10}^{-18}}J\,ato{{m}^{-1}}\] \[\therefore V=\frac{\Delta E}{h}=\frac{2.0437\times {{10}^{-18}}J\,ato{{m}^{-1}}}{6.625\times {{10}^{-34}}\,J\,\,s}\] \[=\text{3}.0\text{84}\times \text{l}{{0}^{\text{15}}}\text{ }{{\text{s}}^{-\text{1}}}\text{ ato}{{\text{m}}^{-\text{1}}}\]
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