Question

An electron is in excited state in a Hydrogen like atom. It has a total energy of -3.4 eV, then

• A. atom may be Lithium
• B. electron may be in third excited state
• C. K.E of electron is 3.4 eV irrespective of atom and its excited state
• D. K.E. of electron is 6.8 eV

Answer 1

Answer: (A), (B), (C)

The correct options are
A atom may be Lithium
B electron may be in third excited state
C K.E of electron is 3.4 eV irrespective of atom and its excited state

$E=-13.6\frac{Z^2}{n^2}\Rightarrow \frac{13.6}{3.4}=\frac{n^2}{Z^2}\Rightarrow \frac{n}{Z}=2$
(i) n = 6 then Z = 3 (Li)
(ii) n = 4 then Z = 2 (He)

And, $K.E=-T.E$