The electron in the hydrogen atom makes a transition from n = 2 energy state to the ground state n = 1. The wavelength of emitted photon is:

Explanation

When an electron in the hydrogen atom transitions from a higher energy state (n = 2) to a lower energy state (n = 1), it emits a photon. The energy of the emitted photon is given by the difference in energy between the two states: ΔE = Efinal - Einitial .The energy levels of the hydrogen atom are given by the formula: En = -Rhc/n2 where R is the Rydberg constant, h is Planck's constant, c is the speed of light, and n is the principal quantum number. For the transition from n = 2 to n = 1, we have ΔE = (-Rhc/12) - (-Rhc/22) = (-Rhc) - (-Rhc/4) = -3Rhc/4 Now, we can use the formula for the wavelength of a photon: λ = h/p = hc/ΔE Substitute the value of ΔE: λ = hc/(-3Rhc/4) = -4/3R The negative sign can be ignored since we are interested in the magnitude of the wavelength, making the correct option: C) 4/3R.