Question

a) How many atoms of Au are in a film that is 1.0 cm square and 25 nm thick? b)What is the weight of the film in a)? c)If the sample is struck by a uniform probe with a diameter of 10 nanometers and penetrates through the film, how many atoms of Au will be in the penetrated volume? Atomic weight Au: 196.96657

Answer 1

Density of Gold, d = g/cm^3 volume o the Gold film, v = A times t where Area, A = cm and inclement t = n m i. e. t = times 10^- 7 cm therefore v = 1 times 25 times 10^- 7 cm^3 Rightarrow v = 25 times 10^- 7 cm^3 Moles weight of Au, M = 196.96657 g Avogadro's number, N_A = 6.022 ti8mes 10^23 therefore number of Atoms in the given film n = N_A/M times v d [Because v d = m i. e. man of the film] n = 6.002 times 10^23/196.96657 times 25 times 10^- 27 times 19.32 n = 1.47671 times 10^17 Weight of film, w = vdg = 25 times 10^- 7 times 19.32 times 980 dyne W = 4.7332 times 10^- 2 dyne or W = 4.7334 times 10^- 5 g w t. Diameter of probe, D = 10 nm = 10^- 6 cm