# To move one electron in one H atom from the ground state to the second excited state, 12.084 eV are needed

How much energy is needed to cause 1 mole of H atoms to undergo this transition

Assume Bohr's model of quantisation

The energy for transition is $$1312\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$

Where the electron moves from $$n_1$$ to $$n_2$$.

$$1312\left(1-\frac19\right)$$ $$1166.22\ \mathrm{kJ/mol}$$

The answer given is 1164. I know we can just account for this by approximation, but I wanna know why is this arising in the first place. Why do I have to approximate? I feel I am doing something wrong, so please help me with my problem.

• Your 1312, whatever units it has, is approximate. – Zhe Sep 13 '19 at 11:48
• Yes it does. But I assumed it was conventionally used. However, after your comment, my concept of reality has completely shattered. – Aditya Sep 13 '19 at 14:07
• Also, notice that the answer you obtained is off by less than 0.2%, so is it really that wrong? – Zhe Sep 14 '19 at 1:20

$$\frac{1}{\lambda} = R_\infty\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=\frac{m_\text{e} e^4}{8 \varepsilon_0^2 h^3 c} \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
As $$\Delta E=\frac{h \cdot c}{\lambda}$$, and as we consider the molar level, therefore:
$$\Delta E_\mathrm{mol} = \frac{N_\mathrm{A} m_\text{e} e^4}{8 \varepsilon_0^2 h^2 } \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
My result for the constant is $$\pu{ 1312.75 kJ/mol}$$ .