A) \[1.54\times 1015\,{{s}^{-}}^{1}\]
B) \[1.03\times 1015\,{{s}^{-}}^{1}\]
C) \[3.08\times 1015\,{{s}^{-}}^{1}\]
D) \[2.0\times 1015\,{{s}^{-}}^{1}\]
Correct Answer: C
Solution :
Ionisation energy of \[H=2.18\times {{10}^{-}}^{18}J\text{ }ato{{m}^{-}}^{1}\] \[\therefore \,\,{{E}_{1}}\] (energy of 1st orbit of H-atom) \[=-2.18\,\,\times \,\,{{10}^{-}}^{18}J\text{ }ato{{m}^{-1}}\] \[\therefore \,\,\,{{E}_{n}}=\frac{-2.18\times {{10}^{-18}}}{{{n}^{2}}}\,\,J\,\,at{{m}^{-\,1}}\] \[Z=\text{1 }for\text{ }H-atom\] \[\Delta E\,\,=\,{{E}_{n}}-{{E}_{1}}=\frac{-2.18\times {{10}^{-}}^{18}}{{{4}^{2}}}\,\,-\,\,\frac{-2.18\times {{10}^{-}}^{18}}{{{l}^{2}}}\] \[\Delta E\,\,=\,\,\,-2.18\times {{10}^{-}}^{18}\,\times \,\frac{-15}{16}\] \[= \,+2.0437 \times 1{{0}^{-}}^{18}J atom{{\,}^{-}}^{1}\] \[\nu =\frac{\Delta E}{h}\,\,=\,\,\frac{2.04\times {{10}^{-18}}}{6.626\times {{10}^{-34}}}\,\,=\,\,3.084\times {{10}^{15}}\,{{s}^{-\,1}}\,atom{{\,}^{-\,1}}\]
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