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 \[r=\frac{{{\varepsilon }_{0}}{{h}^{2}}{{n}^{2}}}{\pi mZ{{e}^{2}}}\] \[\therefore \] \[r\propto \frac{{{n}^{2}}}{Z}\] \[\therefore \] \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{n_{1}^{2}}{n_{2}^{2}}\times \frac{{{Z}_{2}}}{{{Z}_{1}}}\] 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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