JEE MAIN - Chemistry (2019 - 9th April Evening Slot - No. 14)
Which one of the following about an electron
occupying the 1s orbital in a hydrogen atom is
incorrect ?
(The Bohr radius is represented by a0)
(The Bohr radius is represented by a0)
The probability density of finding the
electron is maximum at the nucleus.
The electron can be found at a distance 2a0
from the nucleus
The total energy of the electron is maximum
when it is at a distance a0 from the nucleus.
The magnitude of potential energy is
double that of its kinetic energy on an
average.
Explanation
Probablity density = $${\psi ^2}$$
_9th_April_Evening_Slot_en_14_2.png)
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, En = -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, En = -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.
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, En = -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, En = -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.
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