MCQ
The ionic carbide is
- A$ZnC$
- B$TiC$
- C$SiC$
- ✓$Ca{C_2}$
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$\Psi_{2 s}=\frac{1}{2 \sqrt{2 \pi}}\left(\frac{1}{a_0}\right)^{1 / 2}\left(2-\frac{ r }{ a _0}\right) e ^{- r / 2 a _0}$
At $r=r_0$, radial node is formed. Thus, $r_0$ in terms of $a_0$