A) 0.13
B) 1.06
C) 4.77
D) 2.12
Correct Answer: D
Solution :
| \[r\propto \,n_{1}^{2}\,{{Z}^{2}}\] |
| where n = number of orbit |
| Z = Atomic number |
| \[\because \] \[{{r}_{1}}\propto n_{1}^{2}\] |
| \[{{r}_{2}}\propto \,n_{2}^{2}\,(Z=1\,for\,H-atom)\] |
| So, \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{n_{1}^{2}}{n_{2}^{2}}\] |
| \[\frac{0.530}{{{r}_{2}}}=\frac{{{1}^{2}}}{{{2}^{2}}}\] |
| \[\therefore \] \[{{r}_{2}}=0.530\times 4=2.120\,{\AA}\] |
You need to login to perform this action.
You will be redirected in
3 sec
