UPSC Combined Geo-Scientist, Mains · 2026 · Mains Paper II
Q9. Energy of the J=1 rotational level of H₂
Question
Calculate the energy in joules of the rotational level (with J=1) of the rigid H₂ molecule. Given: r_(H₂)=74.1 pm; relative mass of H atom is 1.0080 amu.
Answer in brief
E_(J=1)=2.42×10⁻²¹ J (0.0151 eV, 60.9×2=122 cm⁻¹ above J=0).
Key points
- μ=1.0080×1.66×10⁻²⁷2=8.366×10⁻²⁸ kg
- I=μ r²=(8.366×10⁻²⁸)(74.1×10⁻¹²)²=4.594×10⁻⁴⁸ kg m²
- (h²)/(8π²I)=(6.626×10⁻³⁴)²8(3.142)²(4.594×10⁻⁴⁸)=1.21×10⁻²¹ J (=B hc, B=60.9 cm⁻¹)
- E_(J=1)=1.21×10⁻²¹×1(2)=2.42×10⁻²¹ J
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
E_J=(h²)/(8π²I)J(J+1), I=μ r²,μ=(m_Hm_H)/(2m_H)=(m_H)/(2)
- μ=1.0080×1.66×10⁻²⁷2=8.366×10⁻²⁸ kg
- I=μ r²=(8.366×10⁻²⁸)(74.1×10⁻¹²)²=4.594×10⁻⁴⁸ kg m²
- (h²)/(8π²I)=(6.626×10⁻³⁴)²8(3.142)²(4.594×10⁻⁴⁸)=1.21×10⁻²¹ J (=B hc, B=60.9 cm⁻¹)
- E_(J=1)=1.21×10⁻²¹×1(2)=2.42×10⁻²¹ J
E₀=0; the J=0→1 line of H₂ would lie at 2B=122 cm⁻¹, in the far infrared; H₂ has no dipole, so it is seen only by rotational Raman (S branch, Δ J=+2).
Coversrigid rotorEnergy of the J=1 rotational level of H₂
All parts of Question 9
- 9 Energy of the J=1 rotational level of H₂10 marks