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JEE Advanced 2016Paper 1
Exam
JEE Advanced
Year
2016
Paper
Paper 1
Subject
Chemistry
Chapter
Structure of Atom

JEE Advanced 2016 · Paper 1 · Chemistry · Structure of Atom

P is the probability of finding the 1s electron of the hydrogen...

PP is the probability of finding the 1s1s electron of the hydrogen atom in a spherical shell of infinitesimal thickness, drdr, at a distance, rr, from the nucleus. The volume of this shell is 4πr2dr4\pi{r}^{2}dr. The qualitative sketch of the dependence of PP on rr is:P is the probability of finding the 1s electron of the hydrogen atom in a spherical shell of infinitesimal thickness,P is the probability of finding the 1s electron of the hydrogen atom in a spherical shell of infinitesimal thickness,P is the probability of finding the 1s electron of the hydrogen atom in a spherical shell of infinitesimal thickness,P is the probability of finding the 1s electron of the hydrogen atom in a spherical shell of infinitesimal thickness,P is the probability of finding the 1s electron of the hydrogen atom in a spherical shell of infinitesimal thickness,

Options

  1. AA long thin hollow metallic cylinder of radius R has a current i ampere. The magnetic induction B away from the axis at
  2. BA long thin hollow metallic cylinder of radius R has a current i ampere. The magnetic induction B away from the axis at
  3. CA long thin hollow metallic cylinder of radius R has a current i ampere. The magnetic induction B away from the axis at
  4. DA long thin hollow metallic cylinder of radius R has a current i ampere. The magnetic induction B away from the axis at

Correct answer

C. A long thin hollow metallic cylinder of radius R has a current i ampere. The magnetic induction B away from the axis at

Step-by-step solution

For 1s1s, the radial part of the wave function is:
Ψ(r)=2(1a0)32e−ra0{Ψ}_{( r )} = 2\left(\dfrac{1}{a0}\right)\dfrac{3}{2}e^{- \dfrac{r}{a0}}
Probability of finding an e–e^{–} in a spherical shell of thickness, ′dr′' dr '~at a distance ′r′' r ' from the nucleus:
P=Ψ(r)2.4\pir2drP = {Ψ}_{\left(r\right)}^{2} . 4\pir^{2}dr
=16\pir2(1a0)3e−2ra0dr= 16\pir^{2}{\left(\dfrac{1}{{a}_{0}}\right)}^{3}e^{\dfrac{- 2r}{{a}_{0}}}dr
So, PP is zero at r=0r = 0 and r=∞r = \infty.
P is the probability of finding the 1s electron of the hydrogen atom in a spherical shell of infinitesimal thickness,
So, the plot of radial probability function, 4\pir2dr.Ψ(r)24\pir^{2}dr . {Ψ}_{\left(r\right)}^{2}~V/s& r\text{r} is as shown above.
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