Chemistry Q50 – NEET-UG 2025

Home Chemistry Structure of AtomQuestion #604

Energy and radius of first Bohr orbit of

{He}^{+}

and

{Li}^{2 +}

are [Given

R_{H} = 2.18 \times 10^{- 18}

a_{0} = 52.9

pm]

Clue 1 of 3: Bohr's Model for Hydrogenic Species

The Bohr model can be applied to single-electron species (hydrogenic atoms/ions) to calculate energy and radius of orbits.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Apply Bohr's Energy and Radius Formulas✦ Active

For hydrogenic species, the energy of the $n$ -th Bohr orbit is given by $E_{n} = - R_{H} \frac{Z^{2}}{n^{2}}$ and the radius by $r_{n} = a_{0} \frac{n^{2}}{Z}$ . Here, $R_{H} = 2.18 \times 10^{- 18}$ J and $a_{0} = 52.9$ pm. For the first Bohr orbit, $n = 1$ .

Step 2: Calculate Energy and Radius for He

^{+}

and Li

^{2 +}

○ Expand

For ${He}^{+}$ : $Z = 2$ , $n = 1$ . \ $E_{1} ({He}^{+}) = - 2.18 \times 10^{- 18} \times \frac{2^{2}}{1^{2}} = - 2.18 \times 10^{- 18} \times 4 = - 8.72 \times 10^{- 18}$ J. \ $r_{1} ({He}^{+}) = 52.9 \times \frac{1^{2}}{2} = 26.45 \approx 26.4$ pm.

For ${Li}^{2 +}$ : $Z = 3$ , $n = 1$ . \ $E_{1} ({Li}^{2 +}) = - 2.18 \times 10^{- 18} \times \frac{3^{2}}{1^{2}} = - 2.18 \times 10^{- 18} \times 9 = - 19.62 \times 10^{- 18}$ J. \ $r_{1} ({Li}^{2 +}) = 52.9 \times \frac{1^{2}}{3} = 17.63 \approx 17.6$ pm.

💡 Teacher's Secret Hint

Remember to use the correct atomic number ( $Z$ ) for each ion.

Step 3: Match Calculated Values with Options○ Expand

Comparing the calculated values with the given options, we find that option (1) matches: \ $E_{n} ({Li}^{2 +}) = - 19.62 \times 10^{- 18}$ J; $r_{n} ({Li}^{2 +}) = 17.6$ pm \ $E_{n} ({He}^{+}) = - 8.72 \times 10^{- 18}$ J; $r_{n} ({He}^{+}) = 26.4$ pm.

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Chemistry Question 50 – NEET-UG 2025

Clue 1 of 3: Bohr's Model for Hydrogenic Species

🧠 Full Solution Path