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Q.

For 2s orbital, the nodal surface exists at the distance

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a

$a_{0}$ from the nucleus

b

$1.5 a_{0}$ from the nucleus

c

$2 a_{0}$ from the nucleus

d

Infinity from the nucleus

answer is C.

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Detailed Solution

At $r = 2 a_{0}$ , three is a node as the value of function $R_{2, 0}$ is zero at this distance.

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For 2s orbital, the nodal surface exists at the distance