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.

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    $\begingroup$ Your 1312, whatever units it has, is approximate. $\endgroup$ – Zhe Sep 13 '19 at 11:48
  • $\begingroup$ Yes it does. But I assumed it was conventionally used. However, after your comment, my concept of reality has completely shattered. $\endgroup$ – Aditya Sep 13 '19 at 14:07
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    $\begingroup$ @Aditya Please try using proper quantity equations and check your units. $\endgroup$ – Faded Giant Sep 13 '19 at 15:11
  • $\begingroup$ Also, notice that the answer you obtained is off by less than 0.2%, so is it really that wrong? $\endgroup$ – Zhe Sep 14 '19 at 1:20

From the Wikipedia page Rydberg constant:

$$\frac{1}{\lambda} = R_\infty\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=\frac{m_\text{e} e^4}{8 \varepsilon_0^2 h^3 c} \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$

As $\Delta E=\frac{h \cdot c}{\lambda}$, and as we consider the molar level, therefore:

$$\Delta E_\mathrm{mol} = \frac{N_\mathrm{A} m_\text{e} e^4}{8 \varepsilon_0^2 h^2 } \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$

Put in the elementary constant values and check the result with your constant.

My result for the constant is $\pu{ 1312.75 kJ/mol}$ .

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