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 of $204eV$ . If it makes a transition to the quantum state $n$ , a photon of energy $40.8eV$ is emitted. Calculate the atomic number $Z$ . Ground state energy of hydrogen atom is $-13.6eV$

Answer 1

Answer: 4

Let ground state energy (in eV) be E₁
Then, from the given condition
$E_{2n}-E_1=204eV$
Or $\frac{{\text{E}}_1}{4{\text{n}}^2}-{\text{E}}_1=\text{204 eV}$
Or $E_1\left(\frac{1}{4n^2}-1\right)=204eV\dots$ (i)
And $E_{2n}-E_n=40.8eV$
Or $\frac{E_1}{4n^2}-\frac{E_1}{n^2}=40.8eV$
$or{E}_1=\left(\frac{-3}{4n^2}\right)=40.8eV\dots .$ (ii)
From Eqs. (i) and (ii),
$\frac{1-\frac{1}{4n}}{\frac{3}{4n^2}}=5$
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 $E_1=-217.6eV$
$E_1=-\left(\begin{array}{l}13.6\end{array}\right)Z_2$
Therefore $Z^2=\frac{E_1}{-13\cdot 6}=\frac{-217-6}{-13\cdot 6}=16$
Therefore $Z=4$