Match List – I with List – II.
|
List – I |
List – II |
|
1. –3.40 |
(I) At infinity |
|
2. –13.6 |
(II) I excited state |
|
3. 0 |
(III) Ground state |
|
4. –0.85 |
(IV) III excited state |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (1) - (II), (2) – (III), (3) – (I), (D) – (IV)
For H-atom (Energy)
$\underset{(n=1)}{\text {Ground state }} E_I=\frac{-13.6}{1^2} eV=-13.6 eV$
$\underset{(n=2)}{I^{st}\text { excited state }} E_{II}=\frac{-13.6}{2^2} eV=-3.4 eV$
$\underset{(n=4)}{{III}^{rd}\text { excited state }} E_{IV}=\frac{-13.6}{4^2} eV=-8.5 eV$
$\underset{(n=\infty)}{\text { For ionisation }} E_{\infty}=\frac{-13.6}{\infty}=0$