Question

A Hydrogen like atom of atomic number Z is in an excited state 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Identity correct statement(s).

• A. Z = 4
• B. n = 4
• C. n = 2
• D. Ground state energy of this atom is -217.6 eV

Answer 1

Answer: (A), (C), (D)

The correct options are
A Z = 4
C n = 2
D Ground state energy of this atom is -217.6 eV

$204=13.6z^2(1-\frac{1}{4n^2})$ ...(1)
$40.8=13.6z^2(\frac{1}{n^2}-\frac{1}{4n^2})$ ...(2)
Dividing (1) by (2)$\Rightarrow 5=\frac{4n^2-1}{3}\Rightarrow n=2$
using (1)
$\frac{204}{13.6}=z^2\frac{15}{16}\Rightarrow z=4$
energy of ground state is,
$E=-13.6z^2=-13.6\times 16=-217.6eV$