Question

The energy level diagram for an hydrogen like atom is shown in the figure. The radius of its first Bohr orbit is
0 eV__ n=$\infty$
-6.04 eV___ n=3
-13.6 eV___ n=2
-54.4 eV___ n=1

• A. $0.265\dot{A}$
• B. $0.53\dot{A}$
• C. $0.132\dot{A}$
• D. $0.142\dot{A}$

Answer 1

The correct option is A $0.265\dot{A}$

We know that $E_n=-13.6\frac{Z^2}{n^2}eV$ and $r_n=0.53\frac{n^2}{Z}\left(\dot{A}\right)$
Here for $n=1,E_1=–54.4eV$
Therefore $-54.4=-13.6\frac{Z^2}{1^1}\Rightarrow Z=2$
Hence radius of first Bohr orbit $r=\frac{0.53(1{)}^2}{2}=0.265\dot{A}$