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 quantum state $n$, a photon of energy $40.8eV$ is emitted. Find $n,Z$ and the ground state energy (in eV) for 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 $-13.6eV$.

• A. $3,4,10.5eV$.
• B. $8,4,10.5eV$.
• C. $2,4,10.5eV$.
• D. $2,4,15.5eV$.

Answer 1

Answer: (C)

$2,4,10.5eV$.

$\text{Given}:$ $E_{2n}-E_1=204eV$, $E_{2n}-E_n=40.8eV$

$\text{Solution}:$

Let us assume that the ground state energy (in eV) be $E_1$

Then from the given condition, we have

$E_{2n}-E_1=204eV$ or

$\frac{E_1}{4n^2}-E_1=204eV$ or

$E_1(\frac{1}{4n^2}-1)=204eV.....\left(i\right)$

$E_{2n}-E_n=40.8eV$ or

$\frac{E_1}{4n^2}-\frac{E_1}{n^2}=40.8eV$ or

$E_1\left(\frac{-3}{4n^2}\right)=40.8eV....\left(ii\right)$

From Eqs. (i) and (ii), we get

$\frac{1-\frac{1}{4n^2}}{\frac{3}{4n^2}}=5$ or

$1=\frac{1}{4n^2}+\frac{15}{4n^2}$ or

$\frac{4}{n^2}=1$ or

n =2

From Eq. (ii)

$E_1=-\frac{4}{3}{n}^2\left(40.8\right)eV=-\frac{4}{3}{n}^2\left(40.8\right)eV$ or

$E_1=-217.6eV{E}_1=-\left(13.6\right)Z^2$

$\therefore Z^2=\frac{E_1}{13.6}=\frac{-217.6}{-13.6}=16$

$\therefore Z=4$

$E_{min}=E_{2n}-E_{2n-1}=\frac{E_1}{4n^2}-\frac{E_1}{(2n-1{)}^2}$

$\therefore E_{min}=10.58eV$

$\text{Hence C is the correct option}$