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 $204\text{eV}$. If it makes a transition to quantum state $n$, a photon of energy $40.8\text{eV}$ is emitted. The value of $n$ will be

• A. $2$
• B. $3$
• C. $1$
• D. $4$

Answer 1

The correct option is A $2$

Let the ground state energy ($\text{in eV}$) be $E_1$
Then from the given condition
maximum energy of photon that can emit is $240\text{eV}$

$\therefore$ $E_{2n}-E_1=204\text{eV}$

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

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

$\text{and}{E}_{2n}-E_n=40.8\text{eV}$

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

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

From equation $\left(i\right)$ and $\left(ii\right),$

$\frac{1-\frac{1}{4n^2}}{\frac{3}{4n^2}}=5\Rightarrow \frac{15}{4n^2}=1-\frac{1}{4n^2}\Rightarrow \frac{4}{n^2}=1$

or, $n=2$

Hence, option $\left(B\right)$ is correct.