In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inner-orbit jumps of the electron for Bohr orbits in an atom of hydrogen?

(A) $3\to2$

(B) $5\to2$

(C) $4\to1$

(D) $2\to5$

Only one option is correct.

My approach

I just drew the electron transition diagram like this: enter image description here

Then, I counted the third line from the red end, i.e., from the left side starting from the first line of the particular series. I then counter checked with the given options. Surprisingly two options matched, i.e., option B as well as C. I don't know further how to decide which is the correct answer out of B and C, since only one answer is correct. There is no other details mentioned in the question and thus I am confused.

  • 1
    $\begingroup$ 4-1 and 5-2 fit the question as you suspect. $\endgroup$
    – porphyrin
    Aug 15, 2019 at 7:32
  • 3
    $\begingroup$ It is kind of an odd question. However since the question mentions a red line, I'd assume that it is referring to the Balmer series which is in the visible range. Thus the answer would be (B). $\endgroup$
    – MaxW
    Aug 15, 2019 at 8:57

1 Answer 1


Since only one answer is correct, we must consider the transitions in the visible region only i.e., the Balmer Series. If multiple answers are correct then it is wise to choose both the options B and C.

The transition $n_2 \to n_1$ can be easily calculated by using the following formula and it is not necessary to draw the transitions, which might be difficult for higher transitions where drawing the transitions would be time-consuming.

$$n_2=n_1 + n$$ where $n$ represents the $n^{th}$ line in a particular series. Since here we are talking about Balmer series, $n_1=2$ and for the third line $n=3$. Substituting the values $$n_2=2+3=5$$

Hence the required transition is given by option B.


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