Atoms And Nuclei Question 265
Question: The energy of $ H{{e}^{+}} $ in the ground state is -54.4 eV, then the energy of $ L{{i}^{++}} $ in the first excited state will be
Options:
A) -30.6 eV
B) 27.2 eV
C) -13.6 eV
D) - 27.2 eV
Show Answer
Answer:
Correct Answer: A
Solution:
-
Energy of electron in nth orbit is
$ E_{n}=-( Rch )\frac{Z^{2}}{n^{2}}=-54.4eV $
For $ H{{e}^{+}} $ is ground state $ E_{1}=-(Rch)\frac{{{(2)}^{2}}}{{{(1)}^{2}}}=-54.4\Rightarrow Rch=13.6 $
$ \therefore $ For $ L{{i}^{++}} $ in first excited state (n=2) $ E=-13.6\times \frac{{{( 3 )}^{2}}}{{{( 2 )}^{2}}}=-30.6eV $
BETA
