Four of the lowest energy levels of $H -$ atom are shown in the figure. The number of possible emission lines should be,

3

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4

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5

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6

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Solution

The correct option is D $6$
Number of possible emission lines will be, $N = \frac{n (n - 1)}{2}$
Here $n = 4$
So, $N = \frac{4 (4 - 1)}{2} = 6$

Hence, $(D)$ is the correct answer.

Why this question?
This equation is based on emission line spectra.

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Four of the lowest energy levels of H−atom are shown in the figure. The number of possible emission lines should be,

The correct option is D 6Number of possible emission lines will be, N=n(n−1)2 Here n=4 So, N=4(4−1)2=6 Hence, (D) is the correct answer. Why this question? This equation is based on emission line spectra.

Four of the lowest energy levels of $H -$ atom are shown in the figure. The number of possible emission lines should be,

The correct option is D $6$
Number of possible emission lines will be, $N = \frac{n (n - 1)}{2}$
Here $n = 4$
So, $N = \frac{4 (4 - 1)}{2} = 6$

Hence, $(D)$ is the correct answer.

Why this question?
This equation is based on emission line spectra.