# Quantization and Bohr's model

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?

• 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. – porphyrin May 20 '18 at 15:52

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 photon has more or less energy than $1.88\ \mathrm{eV}$ (The energy difference needs to be matched exactly!)
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.