You have a metal foil of thickness 200 nm at OK in a vacuum, as shown in the fig

Question

You have a metal foil of thickness 200 nm at OK in a vacuum, as shown in the figure. You instantaneously heat the foil to 1000K. How long does it take for the vacancy concentration to reach 90% of its equilibrium value? The diffusivity of vacancies = D_v = 10^-8 cm^2/s at 1000K. Determine if using only the first term is a valid approximation. Useful relationships: integral_0^L cos(nx)cos(mx)dx =} = 0 for m notequalto n = integral_=L^L cos^2(n pi/L x) dx = x/2 + sin(2 n pi/L x)/4(n pi/L)|_0^L = L for m = n cos(aL) = 0 aL = (2n - 1)pi/2 sin((2n - 1)pi/2) = (1 - 1 + 1 - 1 + 1 ...) = (-1)^n+1

Explanation / Answer

Ans:- as show in figure the metal foli thickness=200 nanometer(nm)

temptarure = 0 kelvien (K)

suppose if our given the instantaneously heat of metal foil= 1000 kelvin (K)

if our vacancy concentration to reach 90% of equilibrium value so much more metal foil particles like electrons, Proton , neutrinos etc have different size known size no actual size known. so this means those wave functions squashed in lot of more minimum that things is initial start to happen related our current laws of physics. that things be like that formation of metal foil in atom electron, proton & neutrinos black holes for vacancy concentration to reach that now not proper developed of gravity in quantum theory that will be needed. so that
the metal foil of will need some kind of quantum spread. so it mean it take equal time of quantum speed so it will vacancy concentration to reach 90 % of equilibrium value.

if diffusivity of vacancies Dv= 10-8cm2/s and temperature 1000 K .as given the relationship metal foil

cos(aL) = 0    aL = (2n-1)3.44 / 2                        it diffusivity first vacancies in 10 -8 cm2/s in intial temp 0 K

Sin ((2n-1) 3.44 / 2) = (1-1 +1 - 1+....) = (-1) n+1 it diffusivity final vacancies in 10 -8 cm2/s in intial temp 1000 K

shows that our accelerated diffusivity vacancies jump rate at 1000 K, at normalized by equilibrium 1000 K value,

E aL (x) = EaL [cos ( (2n-1)/ x.TS0 X 3.44 )+1] / 2.

E aL (x) indicated is characterized by three zero -energy min in all kind of electron ,Proton and neutron ( x = x eq0 ) = 0 and final 1000 K (XaL = vacancy(atom) -site x vac) equilibrium positions, so that other side and one max of energy EaLat the transition state Sin ((2n-1) 3.44 / 2) at the center point of the diffusion-path. The effect of an external energy E (aligned with aL) on the kinetic & potential energy. so that decreases the equilibrium fr diffusivity of vacancies Dv = 10-8 cm2/s at 1000 K. so we determine the monotonically with improve E aL due to some change in X equation. main Start of diffusivity of vacancies from the sinusoidal shape. so that valid approximation useful relationships.

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