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Consider the spectral lines resulting from the transition \(n=2\) to \(n=1\), in the atoms and ions given below, the shortest wavelength is produced by (1) IIydrogen atom (2) Deuterium atom (3) Singly ionised lithium (4) Doubly ionised lithium

Short Answer

Expert verified
Singly ionised lithium or Doubly ionised lithium.

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01

- Understand the Problem

We need to determine which atom or ion produces the shortest wavelength when an electron transitions from the second energy level (=2to the first energy level (=1).
02

- Energy Level Difference

The wavelength of emitted light is inversely proportional to the energy difference between the two levels. This means the larger the energy difference, the shorter the wavelength.
03

- Use the Formula for Energy Levels

The energy levels in a hydrogen-like atom are given by E_n = -13.6Z^2/n^2eV, where n is the principal quantum number and Z is the atomic number. We need to calculate the energy difference for the transition from =2 to =1 for each atom/ion provided.
04

- Calculate Energy Difference for Hydrogen

For Hydrogen atom (Z=1): ΔE = E_1 - E_2 = -13.6(1^2)(1/1^2) - [-13.6(1^2)(1/2^2)] = -10.2 eV.
05

- Calculate Energy Difference for Deuterium

For Deuterium atom (Z=1, similar to Hydrogen but with a neutron, does not affect the energy levels): ΔE = -10.2 eV.
06

- Calculate Energy Difference for Singly Ionised Lithium

For Singly ionised lithium (Z=3): ΔE = -13.6(3^2) (1/1^2) - [-13.6(3^2)(1/2^2)] = -91.8 eV.
07

- Calculate Energy Difference for Doubly Ionised Lithium

For Doubly ionised lithium (Z=3 but missing two electrons): ΔE = -13.6(3^2) (1/1^2) - [-13.6(3^2)(1/2^2)] = -91.8 eV.
08

- Compare Energy Differences

The shorter wavelength will come from the transition with the largest energy difference. Comparing -10.2 eV and -91.8 eV, the largest energy difference corresponds to the Singly ionised and Doubly ionised lithium ions.
09

- Identify the Shortest Wavelength

Between Singly ionised and Doubly ionised lithium, both have same ΔE and hence same wavelength. Therefore, the shortest wavelength is produced by Singly ionised lithium and Doubly ionised lithium.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

energy levels
In atoms, electrons reside in specific regions called energy levels or electron shells. These levels are denoted by quantum numbers such as n=1, n=2, and so on. The energy associated with each level increases as we move further from the nucleus.

When electrons transition between these levels, they either absorb or emit energy in the form of photons. The energy difference between two levels determines the energy of the emitted or absorbed photon, given by the equation: \[ \text{Energy Difference} = E_{n_2} - E_{n_1} \]

Where \[E_n = -13.6 \times Z^2 / n^2 \] (in electron volts, where Z is the atomic number).

Thus, understanding energy levels helps in predicting the wavelengths of the spectral lines generated during these transitions.
wavelength calculation
The wavelength of light emitted during an electronic transition is directly influenced by the energy difference between the initial and final energy levels. According to the Planck-Einstein relation, the energy of the photon is inversely proportional to its wavelength, which is given by: \[ E = \frac{hc}{\text{Wavelength}} \]

Here, \[ h \] is Planck's constant (\[6.626 \times 10^{-34} \] J·s), and \[ c \] is the speed of light (\[3.00 \times 10^8 \] m/s). By rearranging the equation, we can find the wavelength: \[ \text{Wavelength} = \frac{hc}{E} \]

Therefore, a larger energy difference results in a shorter wavelength, and vice versa. This principle is key to interpreting atomic spectra.
hydrogen-like atoms
Hydrogen-like atoms are a class of atoms or ions that contain only one electron, similar to a hydrogen atom. Examples include singly ionized helium (He+) and doubly ionized lithium (Li2+).

The energy levels in these hydrogen-like atoms can be calculated using a modified version of the energy level equation for hydrogen: \[ E_n = -13.6 \times Z^2 / n^2 \]

Here, \[ Z \] is the atomic number. In hydrogen-like atoms, the value of \[ Z \] changes, resulting in different energy levels and differing spectral lines from hydrogen.

In our exercise, we compare hydrogen (\[ Z=1 \]) and different ionized states of lithium (\[ Z=3 \]), which demonstrates that increasing \[ Z \] significantly changes the energy level differences and thus the wavelength of emitted light.
spectral lines
Spectral lines are unique patterns of dark or bright lines seen in the spectrum of a light source, caused by the transition of electrons between energy levels in atoms. When an electron drops from a higher energy level to a lower one, it emits a photon with a specific wavelength. This results in an emission line in the spectrum.

Similarly, absorption lines occur when electrons absorb photons and jump to higher energy levels. The wavelength of these lines provides vital information about the elements present in the light source and their electronic structures.

Spectral lines are a fundamental tool in disciplines such as astronomy, where they are used to identify the compositions and properties of distant stars and galaxies. In our exercise, the spectral lines compare the wavelengths resulting from electron transitions in different atoms and ions, helping us determine which produces the shortest wavelength.

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Most popular questions from this chapter

The wave number of first line in the Balmer series of hydrogen is \(15200 \mathrm{~cm}^{\text {? }}\). The wave number of the first line in the Balmer series of \(\mathrm{Be}^{3+}\) is (1) \(2.43 \times 10^{5} \mathrm{~cm}^{-1}\) (2) \(3.43 \times 10^{5} \mathrm{~cm}^{-1}\) (3) \(4.43 \times 10^{5} \mathrm{~cm}^{-1}\) (4) \(5.43 \times 10^{5} \mathrm{~cm}^{-1}\)

Among the following the wrong statement is (1) \Lambdas the number of orbit increases from the nucleus, the difference between the adjacent energy levels decreases. (2) The angular velocity of the ceectron in the \(n\) th orbit of Bohr's hydrogen atom is inversely proportional to \(n^{3}\). (3) According to Bohrs model of hydrogen atom, the angular momentum of the electron is quantised. (4) The total energy of an atomic electron is more than zero.

When electronic transition occurs from an higher energy to a lower energy state with energy difference equal to expressed in electron volts the wavelength of line emitted is approximately equal to (1) \(\frac{12375}{\Delta E} \wedge\) (2) \(\frac{12375}{\Delta E} \times 10^{-8} \mathrm{~cm}\) (3) \(\frac{12375}{\Delta E} \times 10^{-10} \mathrm{~m}\) (4) Either of these

Which of the following statements is wrong? (1) The rule that limits the maximum number of electrons in an orbital to two is Paulis exclusion principle. (2) IIund's rule states that in a set of degenerate orbitals the electrons distribute with parallel spins. (3) The order of filling of electrons in the orbitals of an atom will be \(5 \mathrm{~s}, 4 \mathrm{p}, 3 \mathrm{~d}, 4 \mathrm{~d}, 4 \mathrm{~s}\). (4) The number of unpaired electrons in the electronic configuration \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{4}\) is 2 .

Of the following transitions in hydrogen atom the one which gives an absorption line of lowest frequency is (1) \(n=1\) to \(n=2\) (2) \(n=3\) to \(n=8\) (3) \(n=2\) to \(n=1\) (4) \(n=8\) to \(n=3\)

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