The 2s-orbital of H-atom has a radial node at 2a0 because $${\psi _{2s}}$$ is proportional to
A. $$\left( {\frac{1}{2} + \frac{r}{{{a_0}}}} \right)$$
B. $$\left( {2 + \frac{r}{{{a_0}}}} \right)$$
C. $$\left( {2 - \frac{r}{{{a_0}}}} \right)$$
D. $$\left( {2 - \frac{r}{{2{a_0}}}} \right)$$
Answer: Option C
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