# 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. May 20, 2018 at 15:52

The energy difference between $n=2$ and $n=3$ in Bohr's model is $$\Delta E=13.6\ \mathrm{eV} \times \left(\frac{1}{4}-\frac{1}{9}\right) = 1.88\ \mathrm{eV} = 3\times10^{-22}\ \mathrm{kJ} = 181.4\ \mathrm{kJ/mol}$$ 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.