What is the maximum number of emission lines when the excited electron of H atom in n 6 drop to the ground state?

A total number of 15 lines (5 + 4 + 3 + 2 + 1) will be obtained in this hydrogen emission spectrum.

The total number of spectral lines emitted when an electron drops to the ground state from the ‘nth‘ level may be computed using the following expression:

$\frac{n(n-1)}{2}$

Since n = 6, total no. spectral lines =

Text Solution

Solution : When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible: Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum. The number of spectral lines produced when an electron in the nth level drops down to the ground state is given by `(n(n-1))/2` given n=6 number of spectral lines ` = (6(6-1))/2 = 15`

What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?

When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible:

Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.

The number of spectral lines produced when an electron in the nth level drops down to the ground state is given by `("n"("n"-1))/2`

Given, n = 6

Number of spectral lines =` (6(6-1))/2 = 15`

Concept: Bohr’s Model for Hydrogen Atom

Is there an error in this question or solution?

(A) 10 lines

(B) 12 lines

(D) 18 lines

The Correct Answer is (C) 15 lines.

Solution:

The maximum number of emission lines when the excited electron of the H atom in n = 6 drops to the ground state is given by:

$\begin{array}{l} \frac{n(n – 1)}{2}\\\Rightarrow \frac{6(6 – 1)}{2}\\\Rightarrow \frac{6(5)}{2}\\\Rightarrow \frac{30}{2}\\\Rightarrow 15. \end{array} $

Therefore, the maximum number of emission lines is 15 lines.

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