1hehe too easy for nodes probability of finding electron is zero
so
\Psi_{2s}=\frac{1}{4\sqrt{2\pi}}\left(\frac{1}{a_0} \right)^{3/2}\left(2-\frac{r_0}{a_0} \right)e^{-r_0/a_0} =0
so 2-r0/a0=0
hence r0=2a0
The Schorodinger wave equation for hydrogen atom is
ψ\Psi_{2s}=\frac{1}{4\sqrt{2\pi}}\left(\frac{1}{a_0} \right)^{3/2}\left(2-\frac{r_0}{a_0} \right)e^{-r_0/a_0}
where a0 is the Bohr's radius, if the radial node is 2s be at r0 then find r0 in terms of a0
Easy dont get anxious.. This is a 2 marks question from JEE 2004
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