| List I | List II | ||
| P. | P.E. of electron in ground state | 1. | \[-54.4\text{eV}\,\text{ato}{{\text{m}}^{-1}}\] |
| Q. | K.E. of electron in ground state | 2. | \[+13.6\,\text{eV}\,\text{ato}{{\text{m}}^{-1}}\] |
| R. | Total energy of in ground state | 3. | \[-108.8\,\text{eV}\,\text{ato}{{\text{m}}^{-1}}\] |
| S. | Ionization energy of He+ in lowest excited state | 4. | \[+54.4\,\text{eV}\,\text{ato}{{\text{m}}^{-1}}\] |
A) \[P-1,Q-2,R-3,S-4\]
B) \[P-4,Q-1,R-2,S-3\]
C) \[P-4,Q-2,R-1,S-3\]
D) \[P-3,Q-4,R-1,S-2\]
Correct Answer: D
Solution :
[d]:\[{{E}_{n}}=-\frac{13.6{{Z}^{2}}}{{{n}^{2}}}eV\,ato{{m}^{-1}}\] (P) For \[H{{e}^{+}},Z=2;\] for ground state of \[H{{e}^{+}},n=1\] Hence, \[{{E}_{1}}=-13.6\times {{2}^{2}}=-54.4eV\,\text{ato}{{\text{m}}^{-1}}\] But\[{{E}_{1}}=P.E.+K.E.\] \[-54.4eV\,ato{{m}^{-1}}=P.E.-\frac{P.E.}{2}=\frac{P.E.}{2}\] \[\Rightarrow \]\[P.E.=-108.8\text{ }eV\text{ ato}{{\text{m}}^{-}}^{1}\] \[(Q)K.E.=-\frac{P.E.}{2}=-\frac{-(-108.8)}{2}=+54.4eV\,\text{ato}{{\text{m}}^{-1}}\]\[\left( R \right){{E}_{1}}={{E}_{total}}=-54.4\text{ }eV\text{ ato}{{\text{m}}^{-}}^{1}\] (S) For ionization of 1st excited state of \[H{{e}^{+}},{{n}_{1}}=2,{{n}_{2}}=\infty \] Hence, \[I.E.=\Delta E=13.6\times {{Z}^{2}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] \[=13.6\times {{2}^{2}}\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{\infty }^{2}}} \right]=+13.6eV\,\text{ato}{{\text{m}}^{-1}}\]
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