UPSC Civil Services (Mains), Chemistry optional · 2026 · Paper I
Q3(c)(i). Energy of one einstein of 800 nm radiation
Question
Calculate the energy of 800 nm electromagnetic radiation per einstein in SI system.
Answer in brief
E=N_Ahc/λ=1.495×10⁵ J per einstein (149.5 kJ per mole of photons).
Key points
- ε=(hc)/(λ)=1.986×10⁻²⁵8.00×10⁻⁷=2.483×10⁻¹⁹ J (1.55 eV)
- 4pt] E=N_A(hc)/(λ)=6.022×10²³×2.483×10⁻¹⁹=1.495×10⁵J einstein⁻¹=149.5 kJ einstein⁻¹
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Given: λ=800 nm=8.00×10⁻⁷ m; hc=1.986×10⁻²⁵ J m; N_A=6.022×10²³mol⁻¹.
- One photon: ε=(hc)/(λ)=1.986×10⁻²⁵8.00×10⁻⁷=2.483×10⁻¹⁹ J (1.55 eV)
- One einstein is one mole of photons: 4pt] E=N_A(hc)/(λ)=6.022×10²³×2.483×10⁻¹⁹=1.495×10⁵J einstein⁻¹=149.5 kJ einstein⁻¹
Reading the answer: E∝1/λ. 800 nm lies beyond the red end of the visible (400–700 nm). The bond enthalpy of H–I is 299 kJ mol⁻¹, twice this value, so 800 nm light cannot dissociate HI; the photolysis in (c)(ii) needs λ≤400 nm.
Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram
All parts of Question 3
- 3(a)(i) Chemical potential from U, H, A and G10 marks
- 3(a)(ii) Cd–Bi simple eutectic: phase diagram and cooling curves10 marks
- 3(a)(iii) How an acetone–dry ice bath works5 marks
- 3(b) Stokes radius and diffusion coefficient of Zn²⁺10 marks
- 3(c)(i) Energy of one einstein of 800 nm radiation5 marks
- 3(c)(ii) Photodecomposition of HI: rate, quantum yield and order10 marks