A) hydrogen
B) deuterium
C) helium
D) lithium
Correct Answer: C
Solution :
Energy of H-like atoms, \[{{E}_{n}}=-\frac{{{Z}^{2}}Rhc}{{{n}^{2}}}=-\frac{{{Z}^{2}}\times 13.6\,eV}{{{n}^{2}}}\] For ground state \[n=1\] \[{{E}_{1}}=-54.4eV\](given) \[\therefore \] \[-54.4\,eV=\frac{{{Z}^{2}}\times 13.6}{{{(1)}^{2}}}eV\] \[\Rightarrow \] \[{{Z}^{2}}=4\] or \[Z=2\] \[Z=2\] is for helium.
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