Atomic Spectra Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. Define Spectroscopy.

Ans. Spectroscopy is the branch of science that deals with the study of the absorption and emission of light and other radiation by matter.

Ques. Define spectral series.

Ans. The spectrum of an element has wavelengths that show definite regularities which can be classified into certain groups called spectral series.

Ques. The intensity of spectral lines depends on

1. Frequency of photon emitted
2. The wavelength of the photon absorbed
3. Frequency of photon absorbed
4. The number of photons of the same frequency absorbed or emitted

Ans. The correct option is d. The number of photons of the same frequency absorbed or emitted.

Explanation: A high-intensity spectral indicates that the transition of electrons occurred in a larger number of atoms. Hence, the intensity or brightness of spectral lines depends upon the number of photons of the same frequency or wavelength emitted or absorbed.

Ques. The wavelength of the spectral line for an electronic transition depends on

1. the nuclear charge of the ion, containing only one electron
2. the velocity of the electron undergoing transition
3. the difference in the energy levels involved in the transition
4. circumference of the orbits in which transition taking place.

Ans. The correct option is c. the difference in the energy levels involved in the transition.

Explanation: The wavelength of the spectral line for an electronic transition is given by the Rydberg formula

\(\frac{1}{\lambda} =Z^2 R (\frac{1}{n_f^2} - \frac{1}{n_i^2})\)

Where

• Z = atomic number
• R = Rydberg constant = 1.097 x 10⁷ m^-1
• n_f = final energy orbit
• n_i = initial energy orbit

Hence, the wavelength of the spectral line depends on the difference in the energy levels involved in the transition.

Ques. Define Excitation potential.

Ans. The potential difference through which an electron of an atom must be accelerated so that it may go from the ground state to the excited state is called the excitation potential of the atom.

Ques. What are the dark lines in a spectrum?

Ans. The dark lines in a spectrum known as absorption lines are the missing specific wavelengths of light in the electromagnetic spectrum. The absorption of light by atoms or molecules in the medium through which the light passes shows some dark lines in a spectrum.

Ques. Why hydrogen atom doesn’t emit X- rays?

1.  Its energy levels are close to each other
2.  Its energy levels are too far apart
3.  It is too small in size
4.  It has a single electron 

Ans. The correct option is d. It has a single electron.

Explanation: The hydrogen atom doesn’t emit X-rays because it has a single electron.

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Short Answers Questions [2 Marks Questions]

Ques. Define atomic spectra.

Ans. Atomic spectra, also known as line spectra are the spectrum of electromagnetic radiation that is emitted or absorbed by the electrons of matter during the transition from a higher energy orbit to a lower energy orbit.

Ques. What are the main differences between line emission spectra and line absorption spectra?

Ans. When atomic gas or vapor is excited, the radiation emitted has a spectrum that contains a specific wavelength, which is known as line emission spectra. It consists of bright lines on a dark background.

On passing white light through atomic gas or vapor, the spectrum of transmitted light shows some dark lines which replace some bright lines seen in the emission spectrum of the gas. This type of spectrum is known as Absorption spectra.

Ques. What are the uses of atomic spectroscopy?

Ans. Various uses of atomic spectroscopy are

• It is used to identify the elements present in a sample.
• It is used to measure the amount of chemical compounds present in a sample.
• It is used in areas like pharmaceutical industries to find the traces of the materials used.
• Atomic spectroscopy is also used in industrial process monitoring.

Ques. Define Ionisation.

Ans. The process of detaching or knocking out an electron from the atom is called ionization. When an electron is completely removed from an atom, there is a deficit of one electron in the atom and it becomes a positive ion. This process is called ionization.

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Long Answers Questions [3 Marks Questions]

Ques. What is the shortest wavelength present in the Paschen series of spectral lines?

Ans. According to the Rydberg formula, we have

\(\frac{1}{\lambda} =Z^2 R (\frac{1}{n_f^2} - \frac{1}{n_i^2})\)

Where

• Z = atomic number
• R = Rydberg constant = 1.097 x 10⁷ m^-1
• n_f = final energy orbit
• n_i = initial energy orbit

For shortest wavelength in the Pashen series of spectral lines

• n_i = ∞
• n_f = 3

On substituting these value in Rydberg formula, we get

\(\frac{1}{\lambda _{min}} =1^2 \times 1.097 \times 10^7(\frac{1}{3^2}- \frac{1}{\infty ^2})\)

⇒ \(\frac{1}{\lambda _{min}} = 1.097 \times 10^7 \times \frac{1}{9}\)

⇒ λ _min = \(\frac{9}{1.097 \times 10^7}\) = 8.204 x 10^-7 m

⇒ λ_min = 820.4 nm

Hence the shortest wavelength present in the Paschen series of spectral lines is 820.4 nm.

Ques. What is the importance of atomic spectra?

Ans. The importance of atomic spectra is 

• It helps to understand the atomic structure of an atom.
• It gives information about the electron configuration within the atom and various energy levels.
• Every element has a unique atomic spectrum, by analyzing the atomic spectrum various elements are identified from a sample.
• Atomic spectra are used to examine the chemical composition of a substance.

Ques. A difference of 3.2 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

Ans. Given the separation of two energy levels in an atom

E = 3.2 eV = 3.2 x 1.6 x 10^-19 J

⇒ E = 5.12 x 10^-19 J

The relationship between energy and frequency is given by

E = hv

⇒ v = E/h

Where

• E = energy
• h = Planck’s constant = 6.62 x 10^-34 Js
• v = frequency of the radiation emitted

On substituting the values, we get

v = (5.12 x 10^-19) / (6.62 x 10^-34) = 7.734 x 10¹⁴ Hz

Hence the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level is 7.734 x 10¹⁴ Hz.

Ques. What are the different types of atomic spectroscopy?

Ans. There are mainly three types of atomic spectroscopy

1. Atomic Absorption Spectroscopy: In this type of spectroscopy, atoms absorb ultraviolet or visible light to transition to higher energy levels from lower energy levels. It measures the amount of absorption of ground-state atoms in the gaseous state. It is commonly used in the detection of metals.
2. Atomic Emission Spectroscopy: In this type of spectroscopy, atoms are excited from the heat of a flame, arc, plasma, or spark to emit light. This spectroscopy used the intensity of light emitted to determine the quantity of an element.
3. Atomic Fluorescence Spectroscopy: In this type of spectroscopy, a beam of light excites the atoms to emit light. Using the fluorometer, the fluorescence from a sample is then analyzed using a fluorometer. This is commonly used to analyze organic compounds.

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Very Long Answers Questions [5 Marks Questions]

Ques. Draw a neat labeled energy level diagram and explain the different series of spectral lines for a hydrogen atom.

Ans. The energy level diagram of a hydrogen atom is shown below

Energy level diagram of a hydrogen atom

The spectral lines emitted as a result of the transition of electrons from the higher energy states to a particular lower energy state form a spectral series.

According to the Rydberg formula

\(\frac{1}{\lambda} =Z^2 R (\frac{1}{n_f^2} - \frac{1}{n_i^2})\)

1. Lyman series: The spectral lines emitted due to the transition of electrons from any higher energy orbit (n_i = 2, 3, 4, 5,....) to the first orbit (n_f = 1) form a spectral series known as the Lyman series. It is mainly observed in the ultraviolet region of electromagnetic waves.
2. Balmer Series: The spectral lines emitted due to the transition of electrons from any higher energy orbit (n_i = 3, 4, 5, 6,....) to the second orbit (n_f = 2) form a spectral series known as the Lyman series. It is mainly observed in the visible region of electromagnetic waves.
3. Pashchen Series: The spectral lines emitted due to the transition of electrons from any higher energy orbit (n_i = 4, 5, 6....) to the third orbit (n_f = 3) form a spectral series known as the Lyman series. It is mainly observed in the infrared region of electromagnetic waves.
4. Brackett Series: The spectral lines emitted due to the transition of electrons from any higher energy orbit (n_i = 5, 6, 7,....) to the fourth orbit (n_f = 4) form a spectral series known as the Lyman series. It is mainly observed in the infrared region of electromagnetic waves.
5. Pfund Series: The spectral lines emitted due to the transition of electrons from any higher energy orbit (n_i = 6, 7, 8....) to the fifth orbit (n_f = 5) form a spectral series known as the Lyman series. It is mainly observed in the infrared region of electromagnetic waves.

Ques. What is the shortest and longest wavelength present in the Lyman series of spectral lines?

Ans. According to the Rydberg formula, we have

\(\frac{1}{\lambda} =Z^2 R (\frac{1}{n_f^2} - \frac{1}{n_i^2})\)

Where

• Z = atomic number
• R = Rydberg constant = 1.097 x 10⁷ m^-1
• n_f = final energy orbit
• n_i = initial energy orbit

For shortest wavelength in the Lyman series of spectral lines

• n_i = ∞
• n_f = 1

On substituting these value in Rydberg formula, we get

\(\frac{1}{\lambda _{min}} = 1^2 \times 1.097 \times 10^7(\frac{1}{1^2}-\frac{1}{\infty^2})\)

⇒ \(\frac{1}{\lambda _{min}} = 1.097 \times 10^7\)

⇒ λ_min = \(\frac{1}{1.097 \times 10^7}\) = 0.911 x 10^-7 m

⇒ λ_min = 91.1 nm

Hence the shortest wavelength present in the Lyman series of spectral lines is 91.1 nm.

For the longest wavelength in the Lyman series of spectral lines

• n_i = 2
• n_f = 1

On substituting these values in the Rydberg formula, we get

\(\frac{1}{\lambda _{max}} = 1^2 \times 1.097 \times 10^7(\frac{1}{1^2}-\frac{1}{2^2})\)

⇒ \(\frac{1}{\lambda _{max}} = 1.097 \times 10^7 \times \frac{3}{4}\)

⇒ λ_max = \(\frac{4}{3 \times 1.097 \times 10^7}\) = 1.216 x 10^-7 m

⇒ λ_max = 121.6 nm

Hence the longest wavelength present in the Lyman series of spectral lines is 121.6 nm.

Ques. What is the shortest and longest wavelength present in the Balmer series of spectral lines?

Ans. According to the Rydberg formula, we have

\(\frac{1}{\lambda} =Z^2 R (\frac{1}{n_f^2} - \frac{1}{n_i^2})\)

Where

• Z = atomic number
• R = Rydberg constant = 1.097 x 10⁷ m^-1
• n_f = final energy orbit
• n_i = initial energy orbit

For shortest wavelength in the Balmer series of spectral lines

• n_i = ∞
• n_f = 2

On substituting these value in Rydberg formula, we get

\(\frac{1}{\lambda _{min}} = 1^2 \times 1.097 \times 10^7(\frac{1}{2^2}-\frac{1}{\infty^2})\)

⇒ \(\frac{1}{\lambda _{min}} = \frac{1.097 \times 10^7}{4}\)

⇒ λ_min = \(\frac{4}{1.097 \times 10^7}\) = 3.646 x 10^-7 m

⇒ λ_min = 364.6 nm

Hence the shortest wavelength present in the Balmer series of spectral lines is 364.6 nm.

For the longest wavelength in the Lyman series of spectral lines

• n_i = 3
• n_f = 2

On substituting these values in the Rydberg formula, we get

\(\frac{1}{\lambda _{max}} = 1^2 \times 1.097 \times 10^7(\frac{1}{2^2}-\frac{1}{3^2})\)

⇒ \(\frac{1}{\lambda _{max}} = 1.097 \times 10^7 \times \frac{5}{36}\)

⇒ λ_max = \(\frac{36}{5 \times 1.097 \times 10^7}\) = 6.563 x 10^-7 m

⇒ λ_max = 656.3 nm

Hence the longest wavelength present in the Balmer series of spectral lines is 656.3 nm.

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