Chemistry · PYQ NEET UG 2025
हिन्दी Energy and radius of first Bohr orbit of \( \text{He}^+ \) and \( \text{Li}^{2+} \) are [Given \( R_H = 2.18 \times 10^{-18} \text{ J}, a_0 = 52.9 \text{ pm} \)]
- \( E_n(\text{Li}^{2+}) = -19.62 \times 10^{-18} \text{ J}; r(\text{Li}^{2+}) = 17.6 \text{ pm}; E_n(\text{He}^+) = -8.72 \times 10^{-18} \text{ J}; r_n(\text{He}^+) = 26.4 \text{ pm} \)
- \( E_n(\text{Li}^{2+}) = -8.72 \times 10^{-18} \text{ J}; r(\text{Li}^{2+}) = 26.4 \text{ pm}; E_n(\text{He}^+) = -19.62 \times 10^{-18} \text{ J}; r_n(\text{He}^+) = 17.6 \text{ pm} \)
- \( E_n(\text{Li}^{2+}) = -19.62 \times 10^{-16} \text{ J}; r(\text{Li}^{2+}) = 17.6 \text{ pm}; E_n(\text{He}^+) = -8.72 \times 10^{-16} \text{ J}; r_n(\text{He}^+) = 26.4 \text{ pm} \)
- \( E_n(\text{Li}^{2+}) = -8.72 \times 10^{-16} \text{ J}; r_n(\text{Li}^{2+}) = 17.6 \text{ pm}; E_n(\text{He}^+) = -19.62 \times 10^{-16} \text{ J}; r_n(\text{He}^+) = 17.6 \text{ pm} \)
Correct answer · Option 1
\( E_n(\text{Li}^{2+}) = -19.62 \times 10^{-18} \text{ J}; r(\text{Li}^{2+}) = 17.6 \text{ pm}; E_n(\text{He}^+) = -8.72 \times 10^{-18} \text{ J}; r_n(\text{He}^+) = 26.4 \text{ pm} \)
Explanation
Using formulas \( E_n = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2} \text{ J} \) and \( r_n = 52.9 \times \frac{n^2}{Z} \text{ pm} \), calculation for \( \text{Li}^{2+} \) (Z=3) and \( \text{He}^+ \) (Z=2) matches option 1.
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