Question

Q5M.8 Consider particles of fine soot 100 nm in diameter, each consisting of roughly 10° carbon atoms (the mass of a carbon atom is about 12.01 u, where 1 u 1.67 x 10 Imagine a beam of such particles moving at 1 mm/s. (a) What would the beam's de Broglie wavelength be? (b) Imagine that we manage to send the beam through two slits 150 nm wide and 300 nm apart. To separate "bright spots" of the soot-particle interference pattern by more than 1 um, about how far from the slits would we have to place the screen for displaying the pattern? (c) What would be the minimum time required to perform the experiment? kg).

Answer 1

a)
mass of the particles,

m = 10^9*12.01*1.67*10^-27

= 2.006*10^-17 kg

v = 1 mm/s

= 1*10^-3 m/s

de Brglie's wavelength, lamda = h/(m*v)

= 6.625*10^-34/(2.006*10^-17*1*10^-3)

= 3.30*10^-14 m <<<<<<<<<-------------Answer

b) d = 300 nm

let R is the distance between the slits and the creen.

distance between two bright spots,

delta_y = lamda*R/d

R = delta_y*d/lamda

= 1*10^-6*300*10^-9/(3.30*10^-14)

= 9.09 m <<<<<<<<---------------Answer

c) minimum time required,

t = R/v

= 9.09/(1*10^-3)

= 9090 s (or) 151.5 min (or) 2.525 hours <<<<<<<<---------------Answer