A) \[n=3\]
B) \[n=4\]
C) \[n=1\]
D) \[n=2\]
Correct Answer: D
Solution :
Radius of orbit of electron in nth excited state of hydrogen \[r=\frac{{{\varepsilon }_{0}}{{h}^{2}}{{n}^{2}}}{\pi \,mZ{{e}^{2}}}\] \[\therefore \] \[r\propto \frac{{{n}^{2}}}{Z}\] ... (1) \[\therefore \] \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{n_{1}^{2}}{n_{2}^{2}}\times \frac{{{Z}_{2}}}{{{Z}_{1}}}\] But \[{{r}_{1}}={{r}_{2}}\] So, \[n_{2}^{2}=n_{1}^{2}\times \frac{{{Z}_{2}}}{{{Z}_{1}}}\] Here: \[{{n}_{1}}=1\] (ground state of hydrogen), \[{{Z}_{1}}=1\] (atomic number of hydrogen), \[{{Z}_{2}}=4\] (atomic number of Beryllium) \[\therefore \] \[n_{2}^{2}={{(1)}^{2}}=\times \frac{4}{1}\] or \[n_{2}^{2}=4\] or \[{{n}_{2}}=2\]
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