Question

The absolute value of ∆E (J) for an atom is equal to Ephoton, the energy of one photon of light emitted for a transition. If Ephoton is 1.63 x 10-18 J, calculate the wavelength of light, in nanometers. (h = 6.626 x 10-34 J sec ; c = 2.998 x 108 m/s; 1 m = 109 nm).

A. 1.22 x 10-43 nm
B. 122 nm
C. 1.22 x 10-7 nm
D. 1.22 x 10-34 nm

Answer 1

Ans. Energy of the photon is given by-

                        E = hv            - equation 1

; where, h = Plank’s constant = 6.626 x 10^-34 Js   ; v = frequency of photon   

Putting the value of v in above equation-

            1.63 x 10^-18 J = (6.626 x 10^-34 Js) x v

            Or, v = 1.63 x 10^-18 J / (6.626 x 10^-34 Js) = 2.460 x 10¹⁵ s^-1

Hence, frequency of the photon = 2.460 x 10¹⁵ s^-1

# The frequency of an electromagnetic radiation or photon is given by the equation-

                        c = v l                      - equation 2

            Where, c = speed of light = 2.998 x 10⁸ m/s

                        v = frequency

Putting the value of wavelength in equation 2-

                        l = c / v

                        Or, l = (2.998 x 10⁸ m s^-1) / 2.460 x 10¹⁵ s^-1 = 1.2187 x 10^-7 m

                        Or, l = 1.2187 x 10^-7 x 10⁹ nm                     ; [1 m = 10⁹ nm]

                        Hence, l = 1.2187 x 10² = 121.87 nm = 122 nm

Hence, wavelength of photon = 122 nm

Correct option – B. 122 nm