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Mumz [18]
3 years ago
5

131_I_53 is an artificially made radioactive isotope of iodine, produced during the operation of nuclear power plants and the de

tonation of nuclear bombs. Exposure to high concentrations of 131 53I represents a serious risk for human health. However, small doses of this radioactive isotope are put to good use in medical studies. 131 53I is widely used in treating thyroid cancer and in diagnosis of abnormal liver and kidney function.
Required:
What is emitted when the radioactive nucleus of 131_I_53 decays to form the stable isotope of xenon 131_Xe_54?
Physics
1 answer:
cricket20 [7]3 years ago
4 0

Answer:

one beta particle

Explanation:

Since the atomic number of 131_I_53 increases by one to 131_Xe_54, we have the release of a beta particle since only a beta particle can cause a change in atomic number without a change in mass number. Since a beta particle is one electron, ₋₁⁰e, our nuclear equation is

¹³¹I₅₃ → ¹³¹X₅₄ + ₋₁⁰e

The atomic and mass numbers on both sides are balanced

mass number 131 = 131 + 0 = 131

atomic number 53 = 54 - 1 = 53

<u>Thus a beta particle is emitted</u>

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3 years ago
When an airplane is flying 200 mph at 5000-ft altitude in a standard atmosphere, the air velocity at a certain point on the wing
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Answer:

P1 = 0 gage

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Explanation:

Given data

Airplane flying = 200 mph = 293.33 ft/s

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Solution

we know density at height 5000-ft is 2.04 × 10^{-3} slug/ft³

so here P1 + \frac{\rho v1^2}{2}  = P2 + \frac{\rho v2^2}{2}

and here

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because P1 = atmospheric pressure

and so here put here value and we get

P1 + \frac{\rho v1^2}{2}  = P2 + \frac{\rho v2^2}{2}

0 + \frac{2.048 \times 10^{-3} \times 293.33^2}{2}  = P2 + \frac{2.048 \times 10^{-3} \times 400.4^2}{2}  

solve it we get

P2 = 87.9 lb/ft³

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the answer is false

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