An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 600 nm . The net energy absorbed by the atom in this process is n × 10 - 4 eV . The value of n is [Assume the atom to be stationary during the absorption and emission process] (Take h = 6 . 6 × 10 - 34 J s and c = 3 × 10 8 m s - 1 ).

Step-by-step solution

The energy of a photon is given by E = h c λ . It is given that λ 1 = 500   nm , λ 2 = 600   nm . The net energy absorbed is Δ E = h c λ 1 - h c λ 2 = h c 10 - 9 ( 1 500 - 1 600 ) = 6 . 6 × 10 - 34 × 3 × 10 8 × 100 500 × 600 × 10 - 9 = 6 . 6 × 3 30 × 10 - 19   J 1   eV = 1 . 6 × 10 - 19   J = 6 . 6 × 3 30 × 1 . 6 e V = 4125 × 10 - 4   eV