xercse 63 In this final exercise you will follow a 1911 determination of Avogadr
ID: 1885877 • Letter: X
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xercse 63 In this final exercise you will follow a 1911 determination of Avogadro's number (the currently establishod IUPAC value is NA 6.02214179x10basod on the rate of cmission of a-particles by radium (Ra, Z-88), and the ralc of production of helium by radium In case you have forgotlen, an a particle is just a bare nucleus of the helium 4 isoloe He) that cames a positive charge of +2 Radium undergoes decay to yidd radon (Rn/1 86) [171: The tale of emission of -particles by radium was determined by Rutherford and Geiger 1181 to be 3.4x10 a-particles per sccond per 1 g of radium. Then, Boltwood and Rutherfond 191 separated some radium salt from its decomposition products and measured the volume of helium produced afler known times. They found that from 192 mg of radium 6.58 mm at STP of belium was produced in 83 days, and 10.38m at STP in 132 days. The half-life of radon) is 383 days (d). STP stands for "andard temperature and pressure". Until 1982, STP was defined as a lemperature of 273.15 K and prossure of exady 1 atm;since 1982, STP is defined as a temperature of 273.15 K and pressure of exactly I bar 10 Pa 120 Boltwood and Rutherord also derived the following formula for the total quantity of helium prodacod (Q): (58) where x is the rale of prodaction of helium by the radium itself,T period of accumulation of helium, and is the decay constant of radon. From X we can obtain the number of moles of helium produced by i g or radíurn per so ond. COmparing this with the a tual mnter of particles cmitlod by 1 gof radium per second we obtain Avogadro's number. Warning! All calculation sleps (a must be performed in your Matkematica nolebook, and all result (a- must be dearly labelled in the Print Tunctions, or the score for thls exercise will be zero: Te (a) evaluate the right hand side (rhs) of equation (58), fCa (b) arrangecquation Q =T.xrVto x = f(Q.TrV. (c) calculate the decay constant of radon, = (dt), (d) determine the value of x the rale of production of helium by 192 mg of radium for T 83 d and Q 6.58m aSTP page 58 of 60 (e) determine the value of x the rale of peoduction of helium by 192 mg of ) using the Math maina function Meané calculate the average value of x ) convent the average valuc of x from to(-):recall that." - 1 mol of gas at STP occupias 22AL IL-10cna × 10cm × 10cm, 1 cm-20mm - 1 d (day) 24 hours1 hour- 60 min 1 min-60s, (h) calculate the rate of production af heliumbyfadium (node tha the value of calculated al the previouas Mep is the rate of production of helium by 192 g of radium) d) since the rade of emission of ' particles byl g of 'adium is 3.4x10a., .. our Occhnically, Rutherfond's) experimcstal Avogadro's nmber, NAup D calculate the percent emor for NA Relative to the caamently established valae; how wll do you think they did in 1911 Answers: (a) 47A 206x102 (h) 5.54 x 10-mExplanation / Answer
a. given equation 58
Q = xT + 3x*integral(1 - e^(-lambda*t))dt
integrating
Q = xT + 3x*[T - 0 + (e^(-lambda*T) - 1)/lambda]
Q = xT + 3x(T*lambda + e^(-lambda*T) - 1)/lambda
b. Q = f(T,x,lambda) has to be rearranged into
x = f(Q,T,lambda)
x = Q*lambda/(T*lambda + 3(T*lambda + e^(-lambda*T) - 1))
x = Q*lambda/(4T*lambda - 3 + 3e^(-T*lambda))
c. given
from 192 mg radium 6.58 mm^3 at STP helium was produced in 83 days
10.38 mm^3 in 132 days
hence
6.58*lambda/(4*83*lambda - 3 + 3e^(-83*lambda)) = 10.38*lambda/(4*132*lambda - 3 + 3e^(-132*lambda))
(4*132*lambda - 3 + 3e^(-132*lambda)) = 1.5775*(4*83*lambda - 3 + 3e^(-83*lambda))
4.27*lambda + 1.7325 + 3(e^(-132*lambda) - 1.5775*e^(-83*lambda)) = 0
plotting on graph we find the solution
lambda = 0.182407152 per day
d. T1 = 83 days
Q1 = 6.58 mm^3
hence
x1 = 0.26304069 mm^3 at stp / day
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