JEE Main · 2022 · Shift-IImediumATOM-076

The energy of one mole of photons of radiation of wavelength 300 nm is: [h=6.6310-34 Js, NA=6.021023 mol-1, c=3108 ms-1]

Structure of Atom · Class 11 · JEE Main Previous Year Question

Question

The energy of one mole of photons of radiation of wavelength 300 nm is:

[h=6.63×10−34h=6.63\times10^{-34} Js, NA=6.02×1023N_A=6.02\times10^{23} mol−1^{-1}, c=3×108c=3\times10^8 ms−1^{-1}]

Options
  1. a

    235 kJ mol−1^{-1}

  2. b

    325 kJ mol−1^{-1}

  3. c

    399 kJ mol−1^{-1}

    ✓
  4. d

    435 kJ mol−1^{-1}

Correct Answerc

399 kJ mol−1^{-1}

Detailed Solution

🧠 Energy by the Batch A mole of photons (NAN_A) turns a tiny microscopic pulse into significant chemical energy.

🗺️ The Macroscopic Calculation

  1. Energy for One: E=hcλ=6.63×10−34×3×108300×10−9=6.63×10−19 JE = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} = 6.63 \times 10^{-19} \text{ J}.
  2. Energy for a Mole: E×6.02×1023≈399,000 J=399 kJ/molE \times 6.02 \times 10^{23} \approx 399,000 \text{ J} = \mathbf{399 \text{ kJ/mol}}.

⚡ The 1200 Shortcut Energy in kJ/mol ≈1.2×105λ(nm)\approx \frac{1.2 \times 10^5}{\lambda (\text{nm})}. For 300 nm300 \text{ nm}, 120,000/300=400120,000/300 = 400.

⚠️ Common Traps Units! JJ vs kJkJ is where most students stumble.

Answer: (C)\boxed{\text{Answer: (C)}}

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