Question

As per Bohr's model, an electron in an uni-electronic species revolves around the nucleus in circular orbits. What is the radius of the 3rd orbit of $H{e}^{+}$ ion?

• A. $2.38\mathring{A}$
• B. $1.19\mathring{A}$
• C. $0.7\mathring{A}$
• D. $4.76\mathring{A}$

Answer 1

The correct option is A $2.38\mathring{A}$

radius of $3^{rd}$ Bohr orbit in $H{e}^{+}$

= $\mathrm{0.52.9}\frac{n^2}{2}{A}^{\circ }$

= $\mathrm{0.52.9}\frac{3^2}{2}$

= $\mathrm{0.52.9}\times 4.5$

= $2.38A^{\circ }$

option A is correct.