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Find the energy of the photon released in the transition from n₁ = 4 to n₂ = 2 for a hydrogen atom. (Note: Use Rydberg Formula)

Find the energy of the photon released in the transition from n₁ = 4 to n₂ = 2 for a hydrogen atom. (Note: Use Rydberg Formula)

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**Problem (2 points):** Find the energy of the photon released in the transition from \( n_1 = 4 \) to \( n_2 = 2 \) for a hydrogen atom. (Note: Use Rydberg Formula) --- This problem involves calculating the energy released when an electron in a hydrogen atom transitions from a higher energy level to a lower one. The Rydberg Formula is relevant for determining the wavelengths of light emitted from such transitions between energy levels.
Transcribed Image Text:**Problem (2 points):** Find the energy of the photon released in the transition from \( n_1 = 4 \) to \( n_2 = 2 \) for a hydrogen atom. (Note: Use Rydberg Formula) --- This problem involves calculating the energy released when an electron in a hydrogen atom transitions from a higher energy level to a lower one. The Rydberg Formula is relevant for determining the wavelengths of light emitted from such transitions between energy levels.
Expert Solution
Concept and Principle:
  • As per Bohr's model when electrons make a transition a photon is absorbed or released. This creates emission and absorption spectra.

 

  • Rydberg formula is used to calculate the wavelength of these photons. It is given by,

λ=1R∞(n12n22n12−n22)

Here R∞ is the Rydberg constant and has a value of 1.097373 × 107 m-1, and n1 and n2 are the states in which transition takes place.

 

  • We know energy carried by a photon is given by,

E=hcλ

Here h is Planck's constant, c is the speed of light, and λ is the wavelength.

 

  • Combining both equations we get the energy released during the transition as,

E=R∞hc(n12−n22n12n22)

Since R∞hc has a constant value this formula will reduce to,

E=(13.6)(n12−n22n12n22) eV

Here n1 and n2 are the states in which transition takes place.

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