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structure of atom 43 Problem 2.6 Solution The energy (E) of a 300 nm photon is Calculate energy of one mole of photons 14 of radiation whose frequency is 5 ×10 given by Hz. Solution Energy (E) of one photon is given by the expression = 6.626 × 10 –1J E = hν The energy of one mole of photons h = 6.626 ×10 –34J s –19 23 –1 14 –1 = 6.626 ×10 J × 6.022 × 10 mol ν = 5×10 s (given) = 3.99 × 10 J mol –1 –34 14 –1 E = (6.626 ×10 J s) × (5 ×10 s ) The minimum energy needed to remove = 3.313 ×10 –19J one mole of electrons from sodium Energy of one mole of photons = (3.99 –1.68) 10 5J mol –1 5 –1 = (3.313 ×10 –1J) × (6.022 × 10 23mol )1 = 2.31 × 10 J mol –1 = 199.51 kJ mol The minimum energy for one electron Problem 2.7 A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second This corresponds to the wavelength by the bulb. = hc E Solution 34 8 1 6.626 10 J s 3.0 10 m s Power of the bulb = 100 watt = 3.84 10 19J = 100 J s–1  = 517 nm Energy of one photon E = hν = hc/λ (This corresponds to green light) ▯34 8 ▯1 6.626 1▯ J s ▯ 10▯m s Problem 2.9 = ▯9 400 ▯10 m The threshold frequency ν 0for a metal is –19 7.0 ×10 14s . Calculate the kinetic energy = 4.969 × 10 J of an electron emitted15h–1 radiation of Number of photons emitted frequency ν =1.0 ×10 s hits the metal. ▯1 100 J s 20 ▯ Solution ▯9 ▯ 2.012 ▯10 s 4.969 ▯10 J According to Einstein’s equation 2 Problem 2.8 Kinetic energy = ½ m ve=h(ν – ν )0 = (6.626 × 10 –34J s) (1.0 × 105 s – 7.0 When electromagnetic radiation of 14 –1 wavelength 300 nm falls on the surface ×10 s ) of sodium, electrons are emitted with a –34 14 –1 5 –1 = (6.626 × 10 J s) (10.0 × 10 s – 7.0 kinetic energy of 1.68 ×10 J mol . What ×10 14s ) is the minimum energy needed to remove –34 14 –1 an electron from sodium? What is the = (6.626 × 10 J s) × (3.0 × 10 s ) –19 maximum wavelength that will cause a = 1.988 × 10 J photoelectron to be emitted? Reprint 2026-27