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Structure of Atom question

2016 · Shift 1 · Q7
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Structure of Atom question

2016 · Shift 1 · Q7

JEE AdvancedChemistryStructure of AtomMCQ+3 / −1
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 4πr2dr4\pi r^2dr4πr2dr. The quantitative ketch of the dependence of P on r is
  1. A
    JEE Advanced 2016 Paper 1 Offline Chemistry - Structure of Atom Question 51 English Option 1
  2. B
    JEE Advanced 2016 Paper 1 Offline Chemistry - Structure of Atom Question 51 English Option 2
  3. C
    JEE Advanced 2016 Paper 1 Offline Chemistry - Structure of Atom Question 51 English Option 3
  4. D
    JEE Advanced 2016 Paper 1 Offline Chemistry - Structure of Atom Question 51 English Option 4
View written solutionFree

Correct answer: D

  1. What is being asked?

We need the graph of the probability of finding the hydrogen atom 1s1s1s electron in a spherical shell of radius rrr and thickness drdrdr.

This probability is:

P(r) dr=4πr2∣ψ1s∣2drP(r)\,dr = 4\pi r^2 |\psi_{1s}|^2 drP(r)dr=4πr2∣ψ1s​∣2dr

So the radial probability density is:

P(r)=4πr2∣ψ1s∣2P(r)=4\pi r^2 |\psi_{1s}|^2P(r)=4πr2∣ψ1s​∣2
  1. Wave function for hydrogen 1s1s1s orbital

For the 1s1s1s orbital,

ψ1s(r)=1πa03e−r/a0\psi_{1s}(r)=\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}ψ1s​(r)=πa03​​1​e−r/a0​

where a0a_0a0​ is the Bohr radius.

Therefore,

∣ψ1s∣2=1πa03e−2r/a0|\psi_{1s}|^2=\frac{1}{\pi a_0^3}e^{-2r/a_0}∣ψ1s​∣2=πa03​1​e−2r/a0​

Substitute into P(r)P(r)P(r):

P(r)=4πr2(1πa03e−2r/a0)P(r)=4\pi r^2\left(\frac{1}{\pi a_0^3}e^{-2r/a_0}\right)P(r)=4πr2(πa03​1​e−2r/a0​) P(r)=4r2a03e−2r/a0P(r)=\frac{4r^2}{a_0^3}e^{-2r/a_0}P(r)=a03​4r2​e−2r/a0​
  1. Analyze the shape of P(r)P(r)P(r)

We have:

P(r)∝r2e−2r/a0P(r) \propto r^2 e^{-2r/a_0}P(r)∝r2e−2r/a0​

Now examine its behavior:

  • At r=0r=0r=0: P(0)=0P(0)=0P(0)=0 So the graph starts from zero.

  • For small rrr, the r2r^2r2 term dominates, so PPP rises from zero.

  • For large rrr, the exponential term e−2r/a0e^{-2r/a_0}e−2r/a0​ dominates, so PPP falls back toward zero.

Thus the graph must:

  • start at zero,
  • rise to a maximum,
  • then decrease asymptotically to zero.
  1. Position of maximum

To find where the maximum occurs, differentiate:

P(r)=4a03r2e−2r/a0P(r)=\frac{4}{a_0^3}r^2 e^{-2r/a_0}P(r)=a03​4​r2e−2r/a0​

Ignoring the constant factor, differentiate r2e−2r/a0r^2 e^{-2r/a_0}r2e−2r/a0​:

ddr(r2e−2r/a0)=e−2r/a0(2r−2r2a0)\frac{d}{dr}\left(r^2 e^{-2r/a_0}\right) = e^{-2r/a_0}\left(2r-\frac{2r^2}{a_0}\right)drd​(r2e−2r/a0​)=e−2r/a0​(2r−a0​2r2​)

Set equal to zero:

e−2r/a0(2r−2r2a0)=0e^{-2r/a_0}\left(2r-\frac{2r^2}{a_0}\right)=0e−2r/a0​(2r−a0​2r2​)=0 2r(1−ra0)=02r\left(1-\frac{r}{a_0}\right)=02r(1−a0​r​)=0

So,

r=0orr=a0r=0 \quad \text{or} \quad r=a_0r=0orr=a0​

Since r=0r=0r=0 gives zero probability, the maximum occurs at

r=a0r=a_0r=a0​
  1. Conclusion about the sketch

The correct sketch is the one that:

  • begins at the origin,
  • increases,
  • reaches a maximum at r=a0r=a_0r=a0​,
  • then decreases gradually toward zero.

This corresponds to Option D.

  1. Comparison with stored correct answer

Stored correct answer: D

My derived answer: D

They agree.

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