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2 years ago

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Q 32.35: The isotope 235U decays by alpha emission with a half-life of 7.0 x 108 y. It also decays (rarely) by spontaneous fissi

on, and if the alpha decay did not occur, its half-life due to spontaneous fission alone would be 3.0 x 1017 y. At what rate do spontaneous fission decays occur in 1.0 g of 235U

1 answer:

Julli [10]2 years ago

5 0

Answer:

rate of fission =5.89*10^3 1\Year

Explanation:

we know that

rate of fission is given as $rate of fission = \frac{0.69}{(T_{1/2})_{fission}} *\frac{ Mass* Avogardo\ number}{Molar\ mass}$

$(T_{1/2})_{fission} = 3*10^{17} y$

mass = 1.0 g

avogardo number = 6.02*10^23

molar mass of isotopes 235U =235

Putting all value to get rate of emission

rate of fission $= \frac{0.69}{3*10^{17}} *\frac{1.0*6.02*10^{23}}{235}$

rate of fission =5.89*10^3 1\Year

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$v=f \lambda$

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Answer:

38.0 m/s east

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In an elastic collision, kinetic energy is conserved.

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(0.311) (30.3)² + (0.0570) (19.2)² = (0.311) v₁² + (0.0570) v₂²

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Solve the system of equations.

0.311 v₁ = 10.52 − 0.0570 v₂

v₁ = 33.82 − 0.1833 v₂

306.5 = 0.311 (33.82 − 0.1833 v₂)² + 0.0570 v₂²

306.5 = 0.311 (1144 − 12.40 v₂ + 0.03360 v₂²) + 0.0570 v₂²

306.5 = 355.7 − 3.856 v₂ + 0.01045 v₂² + 0.0570 v₂²

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Use quadratic formula.

v₂ = [ 3.856 ± √(14.87 − 13.26) ] / 0.1349

v₂ = 19.2 or 38.0

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Using cosine law

$50^2=130^2+120^2-2\times130\times120\cos\beta$

$\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}$

$\beta=\cos^{-1}(0.923)$

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Using formula of force

$F_{130}=ILB\sin\theta$

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We need to calculate the force on 120 cm side

Using formula of force

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The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

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Wavelength = 600 nm

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For constructive fringe

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$x_{2}=\dfrac{3\lambda D}{2d}$

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$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

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