JEE Advanced 2016 · Paper 1 · Chemistry Q19

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ChemistrySingle correct+3 / -1Question 19 of 54

$P$ is the probability of finding the $1s$ electron of hydrogen atom in a spherical shell of infinitesimal thickness, $dr$ , at a distance $r$ from the nucleus. The volume of this shell is $4\pi r^2dr$ . The qualitative sketch of the dependence of $P$ on $r$ is

P against r: largest at r = 0 and decaying monotonically towards zero.

P against r: rises from zero at r = 0 in a semicircular arc and returns to zero at a finite r.

P against r: large at r = 0, dips to a minimum, rises to a maximum, then decays to zero.

P against r: zero at r = 0, rises to a single maximum, then decays smoothly to zero.

Answer and worked solution

Official answer: D

Radial probability. For the 1s orbital, $\psi^2\propto e^{-2r/a_0}$ , so
$P\propto4\pi r^2e^{-2r/a_0}.$

Shape. The $r^2$ factor makes $P=0$ at $r=0$ . The exponential makes it decay at large $r$ . In between it has a single maximum (at $a_0$ ). That is sketch (D).

(D)

Topics:Orbitals and their shapes

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JEE Advanced 2016 · Paper 1, Chemistry question 19