QJEEM19Ch1b3c: JEE Main 2019 Chemistry - Class 11 JEEM19Ch1b3c Solution: Which one of the following about an electron occupying the 1s…

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