A) four times, its ground state radius
B) twice, its ground state radius
C) same as its ground state radius
D) half of its ground state radius
Correct Answer: A
Solution :
Key Idea: The radius of nth Bohrs orbit of hydrogen atom \[{{r}_{n}}=\frac{{{\varepsilon }_{0}}{{n}^{2}}{{h}^{2}}}{\pi m\,{{e}^{2}}}\] As per key idea, \[{{r}_{n}}={{n}^{2}}{{a}_{0}}\] or \[{{r}_{n}}\propto {{n}^{2}}\] For ground state, \[n=1\] For first excited state, \[n=2\] \[\therefore \] \[\frac{{{r}_{2}}}{{{r}_{1}}}={{\left( \frac{2}{1} \right)}^{2}}=4\] or \[{{r}_{2}}=4{{r}_{1}}\] Therefore, radius of first excited state is 4-times than that of ground state radius in H-atom.
You need to login to perform this action.
You will be redirected in
3 sec
