JEE MAIN - Chemistry (2019 - 10th April Morning Slot - No. 3)
The graph between $${\left| \psi \right|^2}$$ and r (radial distance) is shown below. This represents :
_10th_April_Morning_Slot_en_3_2.png)
3s orbital
2s orbital
2p orbital
1s orbital
Explanation
In this $${\left| \psi \right|^2}$$ vs r graph, value of r is zero at only one point. So this orbital must have one radial node.
Here $${\left| \psi \right|^2} \ne 0$$ at r = 0, so orbital should be s orbital. Then $$l$$ = 0.
We know,
No of radial node = n - $$l$$ - 1
$$ \therefore $$ n - $$l$$ - 1 = 1
$$ \Rightarrow $$ n - 0 - 1 = 1
$$ \Rightarrow $$ n = 2
Therefore orbital is 2s.
Here $${\left| \psi \right|^2} \ne 0$$ at r = 0, so orbital should be s orbital. Then $$l$$ = 0.
We know,
No of radial node = n - $$l$$ - 1
$$ \therefore $$ n - $$l$$ - 1 = 1
$$ \Rightarrow $$ n - 0 - 1 = 1
$$ \Rightarrow $$ n = 2
Therefore orbital is 2s.
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