A proton of wavelength \(\lambda_o\) is emitted when an electron in an atom makes a transition from a level of energy 2E. If the electron transits from \(\f...

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The energy of a photon can be calculated using the formula E = hc/λ where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon. When an electron transitions from a higher energy level to a lower energy level, the energy difference is released as a photon. Therefore, the energy of the emitted photon is equal to the difference in energy between the two levels. In this case, the electron transitions from \(\frac{4}{3}E_k\) to E level, which means the energy of the emitted photon is given by: E = (4/3)Ek - E Now, using the formula E = hc/λ, we can find the wavelength of the photon: λ = hc/E Substituting the expression for E: λ = hc/[(4/3)Ek - E] We know that the wavelength of the proton emitted by the electron transition from 2E to E is given by λ0. Therefore, we can write: 2E = hc/λ0 Rearranging for λ0: λ0 = hc/2E Substituting this expression for 2E in the formula for the energy of the emitted photon, we get: λ = hc/[(4/3)(hc/2λ0) - (hc/2λ0)] Simplifying: λ = 3λ0/4 Therefore, the wavelength of the photon that would be emitted is 3/4 of the wavelength of the proton emitted when the electron transitions from 2E to E. Answer: \(\frac{3}{4}\)\(\lambda_o\)