Question

The characteristic spectrum of an atom is observed as
(A) Pure line spectrum
(B) Emission band spectrum
(C) Absorption line spectrum
(D) Absorption band spectrum

Answer 1

Hint: We are given to find the type of spectrum emitted by an atom. We will thus discuss how a characteristic spectrum is produced. We will also discuss the parameters that determine the spectrum type.

Complete step By Step Solution:
The atoms when energized and the electrons in it get excited and after a few moments return back to their ground state and emit the extra energy as light. But it emits light of only certain wavelengths which further corresponds to only certain colors.
Hence, the characteristic spectrum is a series of colored lines with dark spaces in between.

Hence, the correct option is (A).

Additional Information:
The wavenumber of the light emitted by the atoms is specified by the Rydberg’s formula:
$\overline \nu = \dfrac{1}{\lambda } = {R_H}(\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}})$
Where, $\overline \nu $ is the wavenumber, $\lambda $ is the wavelength of the emitted light, ${R_H}$ is known as the Rydberg’s constant having a value of $10973731.6{m^{ - 1}}$, ${n_1}$ is the initial state of the electrons and ${n_2}$ is the final state of the electrons.
Now, different light series are named on the basis of the final state of the electron.
Which means that it depends on the value of ${n_2}$.
Some series are named as given below:
${n_2} = 1$ is named as Lyman Series.
${n_2} = 2$ is named as Balmer Series.
${n_2} = 3$ is named as the Paschen Series. Also named as Bohr Series.
${n_2} = 4$ is named as Brackett Series.
${n_2} = 5$ is named as Pfund Series.
${n_2} = 6$ is named as Humphreys Series.

Note:
Here the spectrum had dark bands bulk with some colors in between which is similar to that of an emission spectrum. But the option contained an emission band spectrum which is not possible as the atoms emit lines. Thus that was not the answer.