Question

If first excitation energy for the H-like(hypothetical)sample is 24 eV,then binding energy in 3rd excited state is:

• A. $2eV$
• B. $32eV$
• C. $49eV$
• D. $5eV$

Answer 1

The correct option is B $32eV$

The energy of an electron in $n^{th}$ orbit $=E_n=-13.6\frac{z^2}{n^2}$ ev/atom

$n=1$, ground state energy $E\left(1\right)=-B.6z^{2}ev⟶\left(1\right)$

$n=2$, energy of $1^{st}$ exicated state $E\left(2\right)=\frac{-13.6}{4}{z}^{2}ev⟶\left(2\right)$

the first exicitation is given by

$E\left(2\right)-E\left(1\right)=13.6z^2(\frac{-1}{4}+1)=13.6z^2\left(3/4\right)=24ev$ (given)

Then, $z^2=\frac{\left(24\right)\left(4/3\right)}{13.6}=\frac{132}{13.6}$

$\therefore$ bind energy $=13.6\left(\frac{32}{13.6}\right)=32ev$

$\therefore$ Binding energy in $3^{rd}$ exicitated state is $32ev$.