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Problems · Problem 2.6

Q.Calculate energy of one mole of photons of radiation whose frequency is 5×10145 \times 10^{14} Hz.

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The energy of one mole of photons is found by first calculating the energy of a single photon using E=hνE = h\nu, then multiplying by Avogadro’s number. The result is approximately 199.5 kJ/mol.

Why this approach works

The energy of a photon is directly proportional to its frequency — that’s the core idea from Planck’s quantum theory. A single photon of frequency ν\nu carries energy E=hνE = h\nu, where hh is Planck’s constant. But the question asks for the energy of one mole of such photons. A mole is a fixed number of particles: 6.022×10236.022 \times 10^{23} of them. So once you know the energy of one photon, scaling up to a mole is just multiplication by Avogadro’s number.

The only subtlety is units. Planck’s constant is usually given in joule-seconds, so the single-photon energy comes out in joules. Multiply by Avogadro’s number and you get joules per mole — a large number. It’s conventional to express this in kilojoules per mole for readability.

Step-by-step solution

1. Recall the Planck-Einstein relation

The energy of a single photon is:

Ephoton=hνE_{\text{photon}} = h \nu

where h=6.626×10−34 J⋅sh = 6.626 \times 10^{-34} \ \text{J·s} and ν=5×1014 Hz\nu = 5 \times 10^{14} \ \text{Hz} (Hz = s−1^{-1}).

2. Calculate the energy of one photon

Ephoton=(6.626×10−34)×(5×1014)E_{\text{photon}} = (6.626 \times 10^{-34}) \times (5 \times 10^{14})

Multiply the coefficients and the powers of ten separately:

Ephoton=(6.626×5)×10−34+14=33.13×10−20 JE_{\text{photon}} = (6.626 \times 5) \times 10^{-34+14} = 33.13 \times 10^{-20} \ \text{J}

Writing in proper scientific notation:

Ephoton=3.313×10−19 JE_{\text{photon}} = 3.313 \times 10^{-19} \ \text{J}

Tip

A quick check: visible light has frequencies around 101410^{14}–101510^{15} Hz, and photon energies in the range 10−1910^{-19} J. This result is right in that ballpark.

3. Scale up to one mole

One mole contains NA=6.022×1023N_A = 6.022 \times 10^{23} photons. So:

Emole=Ephoton×NAE_{\text{mole}} = E_{\text{photon}} \times N_A

Emole=(3.313×10−19)×(6.022×1023)E_{\text{mole}} = (3.313 \times 10^{-19}) \times (6.022 \times 10^{23})

Multiply coefficients: 3.313×6.022≈19.953.313 \times 6.022 \approx 19.95

Multiply powers of ten: 10−19×1023=10410^{-19} \times 10^{23} = 10^{4}

So:

Emole≈19.95×104 J/mol=1.995×105 J/molE_{\text{mole}} \approx 19.95 \times 10^{4} \ \text{J/mol} = 1.995 \times 10^{5} \ \text{J/mol}

4. Convert to kilojoules per mole

Since 1 kJ=1000 J1 \ \text{kJ} = 1000 \ \text{J}:

Emole=1.995×105103 kJ/mol≈199.5 kJ/molE_{\text{mole}} = \frac{1.995 \times 10^{5}}{10^{3}} \ \text{kJ/mol} \approx 199.5 \ \text{kJ/mol}

Watch out

A common mistake is to forget that hh is in J·s and ν\nu in s−1^{-1}, so the product is in joules — not kilojoules. If you skip the final conversion, you’d report 1.995×1051.995 \times 10^5 J/mol, which is numerically correct but not in the conventional unit for molar energies. Always check whether the problem expects J/mol or kJ/mol.

✓Final answer

The energy of one mole of photons is approximately 199.5 kJ/mol\boxed{199.5 \ \text{kJ/mol}}.

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