# 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. Sep 13 '19 at 14:07
– user7951
Sep 13 '19 at 15:11
• 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

For $$1 \mathrm{H}$$ atom $$=12.084 \mathrm{eV}$$
Now, For 1 mole, $$\mathrm{E}=12.084 \times 6.023 \times 10^{23} \mathrm{eV}$$ $$1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$$ So $$\mathrm{E}=12.084 \times 6.023 \times 1.6 \times 10^{-19} \times 10^{23} \mathrm{~J}$$ $$=12.084 \times 6.023 \times 1.6 \times 10^{4} \mathrm{~J}$$ $$=116.45 \times 10^{4} \mathrm{~J}$$ $$=116.45 \times 10 \times 10^{3} \mathrm{~J}$$ $$=116.45 \times 10 \mathrm{~kJ}$$ $$\mathrm{E}=1164.5 \mathrm{~kJ}$$