Probablity density = $${\psi ^2}$$

JEE Main 2019 (Online) 9th April Evening Slot Chemistry - Structure of Atom Question 124 English Explanation

From the graph you can see probablity density is maximum at the nucleus where r = 0.

$$ \therefore $$ Option A is correct.

(B) For hydrogen atom, electron can be found at a any distance distance from the nucleus.

$$ \therefore $$ Option B is correct.

(C) Distance a₀ from the nucleus = First orbital

$$ \therefore $$ n = 1 at first orbital

As Total energy, E_n = -13.6 $$ \times $$ $${{{Z^2}} \over {{n^2}}}$$ eV

$$ \therefore $$ Total energy for hydrogen at n = 1 is,

E₁ = -13.6 $$ \times $$$${{{1^2}} \over {{1^2}}}$$ eV = -13.6 eV

So total energy is minimum at distance a₀ from the nucleus.

$$ \therefore $$ Option C is incorrect.

(D) Total energy, E_n = -13.6 $$ \times $$ $${{{Z^2}} \over {{n^2}}}$$ eV

Kinetic energy, K.E = 13.6 $$ \times $$ $${{{Z^2}} \over {{n^2}}}$$ eV

Potential energy, P.E = -27.2 $$ \times $$ $${{{Z^2}} \over {{n^2}}}$$ eV

$$ \therefore $$ Magnitude of potential energy is double that of its kinetic energy.

$$ \therefore $$ Option D is correct.