Question

A hydrogen like atom (atomic no $Z$) is in a higher excited state of quantum number $n$. This excited energies $10.2eV$ and $17.00eV$ respectively. Alternatively, the atom from the same excited state can make transition to the second excited state by successively emitting $2$ photons of energies $4.25eV$ and $5.95eV$ respectively. Determine the values of $n$ and $Z$. (Ionisation energy of Hydrogen atom $=13.6Ev)$.

• A. $n=5,z=3$.
• B. $n=6,z=4$.
• C. $n=6,z=3$.
• D. $n=6,z=5$.

Answer 1

Answer: (C)

$n=6,z=3$.

For hydrogen like atoms

$E_n=-13.6\times \frac{Z^2}{n^2}$ eV / atom

Given, $E_n–E_2=10.2+17=27.2eV$…(i)

$E_n–E_3=4.24+5.95=10.2eV$

∴ $E_3–E_2=17$

But,$E_3–E_2=(-13.6\times \frac{Z^2}{9})–(-13.6\times \frac{Z^2}{4})$

$=-13.6\times Z^2[\frac{1}{9}–\frac{1}{4}]$

$\Rightarrow Z=3$

Simlarly,

$E_n–E_2=13.6\times \frac{3^2}{n^2}–[-13.6\times \frac{3^2}{2^2}]$

$=-13.6\times [\frac{9}{n^2}–\frac{9}{4}]=-13.6\times 9[4–\frac{n^2}{4n^2}]\dots \left(ii\right)$

From eq. (i) and (ii),

$-13.6\times 9[4–\frac{n^2}{4n^2}]=27.2$

⇒$n^2=\frac{489.6}{13.6}=36\Rightarrow n=6$