First slide

Bohr's atom model

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 eV. If it makes a transition to
quantum state n, a photon of energy 40.8 eV is emitted. The value of n will be

Moderate

A

1
B

2
C

4
D

3

Solution

Let ground state energy (in eV) be $E_{1}$
Then from the given condition
$E_{2 n} - E_{1} = 204 eV \Rightarrow \frac{E_{1}}{4 n^{2}} - E_{1} = 204 eV \Rightarrow E_{1} (\frac{1}{4 n^{2}} - 1) = 204 eV . . . . . . . (i) and E_{2 n} - E_{n} = 40.8 eV \Rightarrow \frac{E_{1}}{4 n^{2}} - \frac{E_{1}}{n^{2}} = E_{1} (- \frac{3}{4 n^{2}}) = 40.8 eV . . . . . . . . (ii) From equations (i) and (ii), \frac{1 - \frac{1}{4 n^{2}}}{\frac{3}{4 n^{2}}} = 5 \Rightarrow n = 2$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App