Question

In: Physics

A plane-parallel monoenergetic beam of 1O^I2 uncharged particles per second is incident perpendicularly on a layer...

A plane-parallel monoenergetic beam of 1O^I2 uncharged particles per second
is incident perpendicularly on a layer of material 0.02 m thick, having a density
p = 11.3 X lo^3 kg/m3. For values of the mass attenuation coefficient m/p =

1X lo^-3 3 X lo^-4 and 1 X lo^-4 m2/kg, calculate the number of primary par-
ticles transmitted in 1 minute. Compare in each case with the approximation

in Eq. (3.5); give percentage errors.

Solutions

Expert Solution

$t = 0.02 m$

$\rho = 11.3 \times 10^{3} kg/m^{3}$

$N_{0}=10^{12}\times 60 min^{-1}$

$N = N_{0}e^{-\mu L} = N_{0}e^{-(m/\rho)(\rho t)}$

(i)

$m/\rho = 1\times 10^{-3} m^{2}/kg$

$N= (10^{12}\times 60)e^{-(10^{-3})(11.3\times 10^{3})(0.02)}$

$N = 4.786\times 10^{13}min^{-1}$

Approximation:

$N_{L}=N_{0}(1-\mu L)= (10^{12}\times 60)(1-(10^{-3}\times 11.3 \times 10^{3}\times 0.02))$

$N_{L}=4.644 \times 10^{13}min^{-1}$

Percentage errors :

$\frac{N-N_{L}}{N}=\frac{4.786-4.644}{4.786} = 0.029$

2.9%

(ii)

$m/\rho = 3\times 10^{-4} m^{2}/kg$

$N= (10^{12}\times 60)e^{-(3\times10^{-4})(11.3\times 10^{3})(0.02)}$

$N = 5.606\times 10^{13}min^{-1}$

Approximation:

$N_{L}=N_{0}(1-\mu L)= (10^{12}\times 60)(1-(3\times 10^{-4}\times 11.3 \times 10^{3}\times 0.02))$

$N_{L}=5.593 \times 10^{13}min^{-1}$

Percentage errors :

$\frac{N-N_{L}}{N}=\frac{5.606-5.593}{5.606} = 0.0023$

0.23%

(iii)

$m/\rho = 1\times 10^{-4} m^{2}/kg$

$N= (10^{12}\times 60)e^{-(10^{-4})(11.3\times 10^{3})(0.02)}$

$N = 5.866\times 10^{13}min^{-1}$

Approximation:

$N_{L}=N_{0}(1-\mu L)= (10^{12}\times 60)(1-(10^{-4}\times 11.3 \times 10^{3}\times 0.02))$

$N_{L}=5.864 \times 10^{13}min^{-1}$

Percentage errors :

$\frac{N-N_{L}}{N}=\frac{5.866-5.864}{5.866} = 3.4\times 10^{-4}$

0.03%

genius_generous answered 1 year ago

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