In the hydrogen atom, the electron makes a transition from the higher orbit ( i ) to a lower orbit ( f ). The ratio of the radius of the orbits in given by r_i : r_f = 16 : 4 . The wavelength of photon emitted due to this transition is _____ nm. (Given Rydberg constant = 1.0973 10^7 /m)

Step-by-step solution

The radius of the n -th orbit in a hydrogen atom is proportional to the square of the principal quantum number n , i.e., r_n n^2 . Given the ratio of the radii of the initial and final orbits: r_i r_f = 16 4 This implies: n_i^2 n_f^2 = 16 4 n_i = 4 and n_f = 2 The wavelength of the emitted photon is given by the Rydberg formula: 1 = R ( 1 n_f^2 - 1 n_i^2 ) Substituting the values: 1 = 1.0973 10^7 ( 1 2^2 - 1 4^2 ) 1 = 1.0973 10^7 ( 1 4 - 1 16 ) 1 = 1.0973 10^7 3 16 = 16 3 1.0973 10^7 m 4.86 10⁻⁷ m = 486 nm Answer: