JEE Challenger
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Qualitative Sketch of Radial Probability Distribution for Hydrogen Atom 1s Electron

Let PP represent the probability of discovering the 1s electron of a hydrogen atom within a thin spherical shell of infinitesimal thickness drdr located at a distance rr from its nucleus. Given that the volume of this shell equals 4πr2dr4\pi r^2 dr, select the qualitative plot illustrating the variation of PP with respect to rr.

Options

A
Option A
B
Option B
C
Option C
D
Option D
Correct

Step-by-Step Solution

To find the qualitative plot illustrating the variation of radial probability PP with respect to distance rr from the nucleus for a 1s electron in a hydrogen atom, we use quantum mechanics principles:

  1. Wavefunction of 1s1\text{s} electron: The normalized radial wavefunction ψ1s(r)\psi_{1s}(r) for the 1s1\text{s} electron of a hydrogen atom is given by: ψ1s(r)=1πa03e−r/a0\psi_{1s}(r) = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0} where a0a_0 is the Bohr radius.

  2. Probability of finding the electron: The probability PP of finding the electron within a thin spherical shell of thickness drdr at a distance rr from the nucleus is the product of the probability density ∣ψ1s(r)∣2|\psi_{1s}(r)|^2 and the volume of the shell dV=4πr2drdV = 4\pi r^2 dr: P=∣ψ1s(r)∣2⋅dV=(1πa03e−2r/a0)(4πr2dr)P = |\psi_{1s}(r)|^2 \cdot dV = \left(\frac{1}{\pi a_0^3} e^{-2r/a_0}\right) (4\pi r^2 dr)

P=4a03r2e−2r/a0drP = \frac{4}{a_0^3} r^2 e^{-2r/a_0} dr

  1. Analyzing the function P(r)P(r):
  • At r=0r = 0: P(0)=0P(0) = 0 This indicates that the probability of finding the electron in a shell of radius zero is zero.

  • As r→∞r \to \infty: lim⁡r→∞P(r)=0\lim_{r \to \infty} P(r) = 0 The probability decreases asymptotically towards zero as rr becomes very large due to the exponential decay factor e−2r/a0e^{-2r/a_0}.

  • Maximum Probability (Most probable radius): To find the value of rr where PP reaches its maximum, we set the derivative of the radial probability density with respect to rr to zero: ddr(r2e−2r/a0)=0\frac{d}{dr} \left( r^2 e^{-2r/a_0} \right) = 0 2re−2r/a0+r2(−2a0)e−2r/a0=02r e^{-2r/a_0} + r^2 \left(-\frac{2}{a_0}\right) e^{-2r/a_0} = 0 2r(1−ra0)e−2r/a0=02r \left(1 - \frac{r}{a_0}\right) e^{-2r/a_0} = 0

    Solving for r≠0r \neq 0: r=a0r = a_0 Thus, the radial probability PP reaches a single maximum peak at r=a0r = a_0 (the Bohr radius).

  1. Conclusion:
  • The graph must start at the origin (0,0)(0,0).
  • It rises to a single maximum peak at r=a0r = a_0.
  • It decays exponentially and approaches zero asymptotically as r→∞r \to \infty without touching the horizontal axis at any finite distance r>0r > 0.

This behavior is correctly represented in plot (D).

Correct Option: D