Question

If $I$ excitation energy for the $H-$ like (hypothetical) sample is $24eV$, then binding energy in $III$ excired state is:

• A. $2eV$
• B. $3eV$
• C. $4eV$
• D. $5eV$

Answer 1

Answer: (A)

$2eV$

$\text{First(I) excitation energy means energy difference between first(ground) state and second state.}$

If ground state energy is $E_1$ then for $n=2$ the energy becomes $E_1/n^2$

Note- As the energy shown above is due to attractive force so it is $negative$ in magnitude.

So $E_2=E_1/2^2=E_1/4$

difference is $E_1-E_2=3E_1/4$

Excitation energy will be $negative$ of the energy difference.

Given that $-3E_1/4=24eV$ so $E_1=-32eV$

$\text{Third(III) excited state means fourth normal state i.e n=4}$

Energy of $4^{th}state$ is $E_1/4^2=E_1/16=-32/16=-2eV$

Binding energy will be positive in nature so answer is $2eV$