Question

A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon o f204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) of this atom. Also calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is-13.6 eV.

Answer 1

Let ground state energy (in eV) be $E_1$
Then, from the given condition
$E_{2n}-E_1=204 eV$ or $\frac{E_1}{4n^2}-E_1=204 eV$
or $E_1\bigg(\frac{1}{4n^2}-1\bigg)=204 eV$...(i)
and $E_{2n}-E_n=40.8 eV$ or $\frac{E_1}{4n^2}-\frac{E_1}{n^2}=40.8 eV$
or $E_1\bigg(\frac{-3}{4n^2}\bigg)=40.8 eV$...(ii)
From Eqs. (i) and (ii),
$\frac{1-\frac{1}{4n^2}}{\frac{3}{4n^2}}=5$ or $1=\frac{1}{4n^2}+\frac{15}{4n^2}$ or $\frac{4}{n^2}=1$ or $n=2$
From Eq. (ii),
$E_1=-\frac{4}{3}n^2(40.8)eV$
$=-\frac{4}{3}(2)^2(40.8)eV$
or $E_1=-217.6eV$
$E_1=-(13.6)Z^2$
$\therefore Z^2=\frac{E_1}{-13.6}=\frac{-217.6}{-13.6}=16$
$\therefore Z=4$
$E_{min}=E_{2n}-E_{2n-1}=\frac{E_1}{4n^2}-\frac{E_1}{(2n-1)^2}$
$=\frac{E_1}{16}-\frac{E_1}9=-\frac{7}{144}E_1=-\bigg(\frac{7}{144}\bigg)(-217.6)eV$
$\therefore E_(min)=10.58eV$