According to quantization it's said that emitted or absorbed energy is quantized.

Then, when it's said in bohr's model an electron changes its orbit (Let's say it goes to a higher energy shell from $n=2$ to $n=3$) The energy difference between the orbits is $100~\mathrm{kJ}$. Now, according to quantization is the $100~\mathrm{kJ}$ itself a quantum. Or the $100~\mathrm{kJ}$ will be absorbed in quantized way?

  • 2
    $\begingroup$ No not the energy you give as this is so large it has to be that for a mole of atoms therefore the quantum of light (radiation) used to excite one atom is this amount divided by Avogadro's number. $\endgroup$
    – porphyrin
    Commented May 20, 2018 at 15:52

1 Answer 1


The energy difference between $n=2$ and $n=3$ in Bohr's model is \begin{equation} \Delta E=13.6\ \mathrm{eV} \times \left(\frac{1}{4}-\frac{1}{9}\right) = 1.88\ \mathrm{eV} = 3\times10^{-22}\ \mathrm{kJ} = 181.4\ \mathrm{kJ/mol} \end{equation} Thus, for this excitation to occur, a single energy quantum (photon) with an energy of $1.88\ \mathrm{eV} = 3\times10^{-22}\ \mathrm{kJ}$ is required.

The excitation does not happen if:

  • the photon has more or less energy than $1.88\ \mathrm{eV}$ (The energy difference needs to be matched exactly!)
  • the same amount energy is distributed over multiple photons (The energy needs to be provided by a single photon!)

As already has been pointed out, the $181.4\ \mathrm{kJ/mol}$ itself is not a quantum, but one mole of quanta, where each quantum has $1.88\ \mathrm{eV} = 3\times10^{-22}\ \mathrm{kJ}$ of energy. As each atom can only absorb this energy once (assuming it does not relax back to $n=2$ by emitting the energy), to absorb so much energy will also require one mole of atoms.


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