Answer:
To determine which transition between electron energy levels emits a photon with the maximum frequency, we use the relationship between photon frequency and the energy difference between the energy levels involved in the transition.The frequency ( \nu ) of the photon emitted during the transition between two energy levels is given by:[ \nu = \frac{E_{n2} - E_{n1}}{h} ]where:( E_{n1} ) and ( E_{n2} ) are the energies of the initial and final states of the electron,( h ) is Planck's constant.The energy ( E_n ) of an electron in the nth energy level of a hydrogen-like atom (like an electron in an atom) is given by:[ E_n = -\frac{Z^2 \cdot R_H}{n^2} ]where:( Z ) is the atomic number (for hydrogen, ( Z = 1 )),( R_H ) is the Rydberg constant for hydrogen.The maximum frequency of the emitted photon occurs when the transition results in the largest energy difference ( E_{n2} - E_{n1} ). This corresponds to the transition between energy levels where ( n1 ) is higher and ( n2 ) is lower.Let's analyze each option:a) ( n1 = 1 ) to ( n2 = 2 ): [ E_{n1} = -\frac{1}{1^2} = -1 ] [ E_{n2} = -\frac{1}{2^2} = -\frac{1}{4} ] [ E_{n2} - E_{n1} = -\frac{1}{4} - (-1) = \frac{3}{4} ]b) ( n1 = 2 ) to ( n2 = 1 ): [ E_{n1} = -\frac{1}{2^2} = -\frac{1}{4} ] [ E_{n2} = -\frac{1}{1^2} = -1 ] [ E_{n2} - E_{n1} = -1 - (-\frac{1}{4}) = -\frac{3}{4} ]c) ( n1 = 2 ) to ( n2 = 6 ): [ E_{n1} = -\frac{1}{2^2} = -\frac{1}{4} ] [ E_{n2} = -\frac{1}{6^2} = -\frac{1}{36} ] [ E_{n2} - E_{n1} = -\frac{1}{36} - (-\frac{1}{4}) = -\frac{8}{36} + \frac{9}{36} = \frac{1}{36} ]d) ( n1 = 6 ) to ( n2 = 2 ): [ E_{n1} = -\frac{1}{6^2} = -\frac{1}{36} ] [ E_{n2} = -\frac{1}{2^2} = -\frac{1}{4} ] [ E_{n2} - E_{n1} = -\frac{1}{4} - (-\frac{1}{36}) = -\frac{9}{36} + \frac{1}{36} = -\frac{8}{36} = -\frac{2}{9} ]From the calculations, the transition that results in the largest energy difference (and thus the maximum frequency of emitted photon) is option:b) ( n1 = 2 ) to ( n2 = 1 )Therefore, the correct answer is option b).
