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Animated Solution for Physics - Atoms and Nuclei: Consider the spectral line resulting from the transition in the atoms and ions given below. The shortest wavelength is produced by

Select Answer:

Visualized Solution

  • Transition:

  • Energy is released as a photon.

  • Hydrogen ():
  • Deuterium ():

  • Singly ionised Helium ():
  • Doubly ionised Lithium ():

  • To minimize , we must maximize .

  • Maximum for
  • Shortest wavelength is produced by doubly ionised lithium.

  • Maximum Maximum Energy Minimum Wavelength

The Sigma Insight: Bohr's Atomic Model and Energy Levels

Solution Diagram

Finding the Shortest Wavelength in Atomic Transitions

Imagine an electron in a hydrogen-like atom. It is currently in the first excited state, where . It is about to make a jump down to the ground state, where . When this electron drops to a lower energy level, it loses energy. This lost energy is emitted in the form of a photon. The energy of this photon is inversely proportional to its wavelength.

The Rydberg Formula

To find the exact wavelength of this emitted photon, we use the Rydberg formula. This formula connects the wavelength to the atomic number and the principal quantum numbers of the transition:
Let us plug in the values for our specific transition. The electron falls to the ground state, so . It falls from the first excited state, so . Now, let us simplify the terms inside the bracket. One minus one-fourth gives us three-fourths. So, one over lambda equals three over four, times :

Wavelength vs Atomic Number

Rearranging this, we can see that the wavelength is inversely proportional to the square of the atomic number .
This is the crucial relationship we need to solve the problem. We are looking for the shortest wavelength. According to our inverse relationship, to minimize the wavelength, we need to find the atom or ion with the maximum atomic number.

Evaluating the Options

Let us evaluate our options.
Hydrogen () has an atomic number of .
Deuterium () is just an isotope of hydrogen, so it also has an atomic number of .
Next, we have singly ionised helium (). Helium has two protons, so its .
Finally, doubly ionised lithium () has three protons, giving it a .

Conclusion

Comparing the values, doubly ionised lithium has the highest atomic number, which is three. Therefore, it will produce the shortest wavelength for this transition.
You can also think about this in terms of energy. The energy gap between levels scales with . A larger means a larger energy gap, which corresponds to a higher energy photon, and thus, a shorter wavelength.

Similar Questions

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Hydrogen (), deuterium (), singly ionised helium ( and doubly ionised lithium ( all have one electron around the nucleus. Consider an electron transition form to . If the wavelengths of emitted radiation are and respectively for four elements, then approximately which one of the following is correct?

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