A) \[-\,273K\]
B) \[300\,K\]
C) \[1430\,K\]
D) \[14300\,K\]
Correct Answer: D
Solution :
| Given ratio of number of atoms m state 2 to number of atoms in state 1 is 0.001. | |
| So, \[\frac{n\,({{E}_{2}})}{n\,({{E}_{1}})}=\frac{1}{1000}\] | |
| \[\Rightarrow \] \[\frac{1}{1000}=\frac{{{N}_{2}}}{{{N}_{1}}}\cdot ({{e}^{-({{E}_{2}}-{{E}_{1}})/kT}})\] | ?(i) |
| where,\[{{N}_{2}}\]and \[{{N}_{1}}\]are number of possible electron states in \[n=2\] and n =1 quantum states. | |
| Also, \[{{E}_{2}}\] and \[{{E}_{1}}\] are energies in \[n=2\] and \[n=1\] quantum states. |
| Here, \[{{N}_{2}}=8,\]\[{{N}_{1}}=2,\]\[{{E}_{2}}=\frac{{{E}_{1}}}{4}\]and |
| \[{{E}_{1}}=-13.6eV\] |
| \[k=8.62\times {{10}^{-\,5}}\frac{eV}{K}\] |
| Taking log in Eq. (i), we have |
| \[T=\frac{1}{k}\cdot \frac{\frac{3}{4}(-{{E}_{1}})}{\log \,(4000)}\] |
| \[=\frac{\frac{3}{4}\times 13.6}{8.62\times {{10}^{-5}}\times \log 4000}\] |
| \[=1.43\times {{10}^{4}}K\] |
| \[=14300K\] |
You need to login to perform this action.
You will be redirected in
3 sec
