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 total energy of the electron is maximum when it is at a distance a0 from the nucleus.

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 magnitude of potential energy is double that of its kinetic energy on an average.

Structure of Atom PYQ — JEE Main 2019 · 9 Apr · Shift 2 · Q14 with Solution

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Correct answer: C

For a hydrogen atom in the $1s$ orbital, the wave function is
$\psi_{1s}(r)=\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}$
Hence the probability density is
$|\psi_{1s}(r)|^2=\frac{1}{\pi a_0^3}e^{-2r/a_0}$
Check option A: The probability density $|\psi_{1s}(r)|^2$ is maximum at $r=0$ , i.e. at the nucleus.

So, A is correct.
Check option B: In quantum mechanics, the electron is not confined to a fixed radius. For the $1s$ orbital, the wave function is nonzero for all finite $r$ . Thus the electron can be found at $r=2a_0$ as well.

So, B is correct.
Check option D: For a hydrogen atom, by the virial theorem:
$\langle V \rangle = -2\langle T \rangle$
Therefore,
$|\langle V \rangle|=2\langle T \rangle$
So the magnitude of potential energy is double the kinetic energy on average.

Hence, D is correct.
Check option C: In the hydrogen atom, the electron in the $1s$ state has a definite total energy:
$E=-13.6\,\text{eV}$
This is the energy eigenvalue of the state and does not depend on the electron being at a particular distance such as $r=a_0$ .

Therefore the statement "the total energy of the electron is maximum when it is at a distance $a_0$ from the nucleus" is incorrect.
Conclusion: The incorrect statement is option C.

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Correct answer: C

For a hydrogen atom in the $1s$ orbital, the wave function is
$\psi_{1s}(r)=\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}$
Hence the probability density is
$|\psi_{1s}(r)|^2=\frac{1}{\pi a_0^3}e^{-2r/a_0}$
Check option A: The probability density $|\psi_{1s}(r)|^2$ is maximum at $r=0$ , i.e. at the nucleus.

So, A is correct.
Check option B: In quantum mechanics, the electron is not confined to a fixed radius. For the $1s$ orbital, the wave function is nonzero for all finite $r$ . Thus the electron can be found at $r=2a_0$ as well.

So, B is correct.
Check option D: For a hydrogen atom, by the virial theorem:
$\langle V \rangle = -2\langle T \rangle$
Therefore,
$|\langle V \rangle|=2\langle T \rangle$
So the magnitude of potential energy is double the kinetic energy on average.

Hence, D is correct.
Check option C: In the hydrogen atom, the electron in the $1s$ state has a definite total energy:
$E=-13.6\,\text{eV}$
This is the energy eigenvalue of the state and does not depend on the electron being at a particular distance such as $r=a_0$ .

Therefore the statement "the total energy of the electron is maximum when it is at a distance $a_0$ from the nucleus" is incorrect.
Conclusion: The incorrect statement is option C.