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Mrrafil [7]
3 years ago
7

During nuclear decay, a new isotope is created. How is the nucleus of the new isotope different from the parent if the parent is

otope has undergone alpha decay?
Physics
2 answers:
Veseljchak [2.6K]3 years ago
6 0

Answer:

the difference in parent nuclei and daughter nuclei is that daughter nuclei is having smaller mass number and by 4 and smaller atomic number by 2

_z^AX ----> _{z-2}^{A-4}Y + _2^4He

Explanation:

Alpha nucleus is the helium nuclei which is having 2 protons and 2 neutrons

So we will have

\alpha = _2^4He

now we know that in all radioactive decay mass number and atomic number will remain constant on product and reactant side

So we have

X ------> Y + \alpha

now by mass number conservation and atomic number conservation we have

_z^AX ----> _{z-2}^{A-4}Y + _2^4He

so the difference in parent nuclei and daughter nuclei is that daughter nuclei is having smaller mass number and by 4 and smaller atomic number by 2

STALIN [3.7K]3 years ago
3 0

Answer:

Alpha decay will produce a daughter nucleus with more protons and beta decay will produce a daughter nucleus with fewer protons than the parent nucleus has.

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From coulomb's law,

Formula:

F = kqq'/r²........................ Equation 1

Where F = Force of repulsion, k = coulomb's constant, q = first positive charge, q' = second positive charge, r = distance between the charge.

Given: q = 20 μC = 20×10⁻⁶ C, q' = 100 μC = 100×10⁻⁶ C, r = 150 cm = 1.5 m.

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Substitute these values into equation 1

F = (20×10⁻⁶ )( 100×10⁻⁶)(9×10⁹)/1.5²

F = 1800×10⁻³/2.25

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A 50 Kg box sits at rest on a 30 degree ramp where the coef of static friction is 0.5773. If your push was directed at an angle
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<u>Given data:</u>

m= 50 Kg,

W= m×g = 50 × 9.81 = 490.5 N

ramp angle (α) = 30 degrees,

coefficient of friction (μs) = 0.5773,

Push at an angle (Θ) = 40 degrees,

Determine: Push to get box move up (P)=?

From the figure,

Resolving the forces along the plane

W sinα + μs.R = P cos Θ       --------------------- (i)

Resolving the forces perpendicular to the inclined plane

W cosα = R+Psin Θ  =>  R= W cosα - Psin Θ -------------- (ii)

Solving (i) and (ii) and keeping <em>μs = tan Φ, Φ = Θ </em>

<em>Pmin = W sin( α +Θ  )</em>

<em>          = W[ sin α.Cos Θ + cos α.sin Θ]</em>

<em>           = 490.5 [ (sin 30.cos40) + (cos30.sin 40)]</em>

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