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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 217 E7A.5(b) �e energy of a photon with wavelength λ is given by E = hν = hc/λ = (6.6261 × 10−34 J s) × (2.9979 × 108ms−1)/λ = (1.9825 × 10−25 J)/(λ/m) �e energy per mole is given by Em = NAE = (0.11939 J mol−1)/(λ/m) Hence, the following table is drawn up λ E Em/kJmol−1 (a) 200 nm 991 zJ 597 (b) 150 pm 1.32 pJ 7.96 × 105 (c) 1.00 cm 19.8 yJ 1.19 × 10−2 E7A.6(b) When a photon is absorbed by a free hydrogen atom, the law of conservation of energy requires the kinetic energy acquired by the atom is Ek, the energy of the absorbed photon. Assuming relativistic corrections are negligible the kinetic energy is Ek = Ephoton = 1 2mHυ2.�e atom is accelerated to the speed, υ = ( 2Ephoton mH ) 1/2 = ( 2NAEphoton MH ) 1/2 = ( 2 × (6.0221 × 1023mol−1) × Ephoton (1.0079 × 10−3 kgmol−1) ) 1/2 = (3.45... × 1013 ms−1) × (Ephoton/J) 1/2 �e photon energies have been calculated in Exercise E7A.5(b), and thus the following table can be drawn up λ E υ/km s−1 (a) 200 nm 991 zJ 17.3 (b) 150 pm 1.32 pJ 631 (c) 1.00 cm 19.8 yJ 0.0773 E7A.7(b) �e total energy emitted from a laser at (constant) power P in a time interval ∆t is P∆t.�e energy of a single photon of wavelength λ is E = hc/λ. Hence, the total number of photons emitted in this time interval is the total energy emitted divided by the energy per photon (assuming the light is monochromatic) N = P∆t Ephoton = P∆tλ hc �us, for a time interval of 1 s and a wavelength of 700 nm
