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QJEEAdv16Chb70b: JEE Advanced 2016 Chemistry - Class 11 JEEAdv16Chb70b Solution: P is the probability of finding the 1s electron of hydrogen…

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^2dr . The quantitative ketch of the dependence of P on r is

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/042fbd7662547ba042c6e570fdccc49431745780ec7a078d3b134d77afb8e59d.webp
  • B https://ik.imagekit.io/novatra/photos/370207979ad80a1cec850dde195dfa14882d01523cb23616fcf470e758fe674a.webp
  • C https://ik.imagekit.io/novatra/photos/e8e80bbe5941dcff04a923cff66a49c8e69fe6d724b2552582e83b5b32c7d502.webp
  • D https://ik.imagekit.io/novatra/photos/4aefb10b21d2d690762a5954490c785f3b0f4b5aa8651f0139e96024e43b082a.webp

Step-by-Step Verified Solution

The probability of finding an electron of hydrogen atom in a spherical shell of infinitesimal thickness, drd r , at a distance rr from the nucleus, with volume dV=4πr2drd V=4 \pi r^2 d r is

P=ψ2⋅dV=ψ2⋅4πr2dr=R2(r)4πr2drP=\psi^2 \cdot d V=\psi^2 \cdot 4 \pi r^2 d r=R^2(r) 4 \pi r^2 d r

Here, R2(r)R^2(r) is the radial density function.

For 1s1 s subshell, n=1,l=0n=1, l=0 and n−l−1=0n-l-1=0 . Thus, the number of radial and angular nodes =0=0 . For 1s orbital, the probability of finding an electron will be maximum at near to the the nucleus, as 1s orbital is the nearest to the nucleus as it is the lowest orbital in terms of energy. Therefore, the plot of radial probability P=R2(r)4πr2P=R^2(r) 4 \pi r^2 versus rr is as follows :

JEE Advanced 2016 Paper 1 Offline Chemistry - Structure of Atom Question 51 English Explanation

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