JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

Which of the following statements regarding the energy of the stationary state is true in the following one - electron systems ?

Official previous-year question

Held on 23 Jan 2026 · Verified 6 Jul 2026.

Options

A
$+2.18 \times 10^{-18} \mathrm{~J} \text { for second orbit of } \mathrm{He}^{+} \text {ion }$
B
$-2.18 \times 10^{-18} \mathrm{~J} \text { for third orbit of } \mathrm{Li}^{2+} \text { ion }$
C
$-1.09 \times 10^{-18} \mathrm{~J} \text { for second orbit of } \mathrm{H} \text { atom. }$
D
$+8.72 \times 10^{-18} \mathrm{~J} \text { for first orbit of } \mathrm{He}^{+} \text {ion }$

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Solution

The energy of an electron in the $n^{th}$ stationary state of a one-electron system is given by the Bohr's formula:

$E_n = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2}$ J

where $Z$ is the atomic number and $n$ is the principal quantum number.

For option (1): Second orbit of $He^+$ ( $Z=2, n=2$ )

$E = -2.18 \times 10^{-18} \times \frac{2^2}{2^2} = -2.18 \times 10^{-18}$ J. (Incorrect sign in option)

For option (2): Third orbit of $Li^{2+}$ ( $Z=3, n=3$ )

$E = -2.18 \times 10^{-18} \times \frac{3^2}{3^2} = -2.18 \times 10^{-18}$ J. (Correct)

For option (3): Second orbit of $H$ atom ( $Z=1, n=2$ )

$E = -2.18 \times 10^{-18} \times \frac{1^2}{2^2} = -0.545 \times 10^{-18}$ J. (Incorrect value)

For option (4): First orbit of $He^+$ ( $Z=2, n=1$ )

$E = -2.18 \times 10^{-18} \times \frac{2^2}{1^2} = -8.72 \times 10^{-18}$ J. (Incorrect sign in option)

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Options

$+2.18 \times 10^{-18} \mathrm{~J} \text { for second orbit of } \mathrm{He}^{+} \text {ion }$

$-2.18 \times 10^{-18} \mathrm{~J} \text { for third orbit of } \mathrm{Li}^{2+} \text { ion }$

$-1.09 \times 10^{-18} \mathrm{~J} \text { for second orbit of } \mathrm{H} \text { atom. }$

$+8.72 \times 10^{-18} \mathrm{~J} \text { for first orbit of } \mathrm{He}^{+} \text {ion }$

Solution

The energy of an electron in the $n^{th}$ stationary state of a one-electron system is given by the Bohr's formula:

$E_n = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2}$ J

where $Z$ is the atomic number and $n$ is the principal quantum number.

For option (1): Second orbit of $He^+$ ( $Z=2, n=2$ )

$E = -2.18 \times 10^{-18} \times \frac{2^2}{2^2} = -2.18 \times 10^{-18}$ J. (Incorrect sign in option)

For option (2): Third orbit of $Li^{2+}$ ( $Z=3, n=3$ )

$E = -2.18 \times 10^{-18} \times \frac{3^2}{3^2} = -2.18 \times 10^{-18}$ J. (Correct)

For option (3): Second orbit of $H$ atom ( $Z=1, n=2$ )

$E = -2.18 \times 10^{-18} \times \frac{1^2}{2^2} = -0.545 \times 10^{-18}$ J. (Incorrect value)

For option (4): First orbit of $He^+$ ( $Z=2, n=1$ )

$E = -2.18 \times 10^{-18} \times \frac{2^2}{1^2} = -8.72 \times 10^{-18}$ J. (Incorrect sign in option)

JEE Main 2026 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry