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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. $\endgroup$
    – user7951
    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
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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 $\mathrm{E}=12.084 \times 6.023 \times 1.6 \times 10^{-19} \times 10^{23} \mathrm{~J}$ $$ =12.084 \times 6.023 \times 1.6 \times 10^{4} \mathrm{~J} $$ $$ =116.45 \times 10^{4} \mathrm{~J} $$ $$ =116.45 \times 10 \times 10^{3} \mathrm{~J} $$ $$ =116.45 \times 10 \mathrm{~kJ} $$ $$ \mathrm{E}=1164.5 \mathrm{~kJ} $$

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Slight differences of calculated and expected numerical results are matter of different input data, different rounding and sometimes different computation procedure or even an error.

Eventual acceptation of slight deviations or insisting on the exact match is rather matter of the local education policy, regardless of which is correct.

Chemistry itself has little to do with that.

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