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Harman [31]
3 years ago
12

Can alpha radiation affect electronics

Chemistry
1 answer:
Tom [10]3 years ago
5 0

Answer:

Radiation effects on electrical equipment depend on the equipment and on the type of ionizing radiation to which it is exposed.

 

First, beta radiation has little, if any, effect on electrical equipment because this type of ionizing radiation is easily shielded. The equipment housing and the construction of the parts within the housing will protect the equipment from beta-radiation (high-energy electrons) exposure.

 

Gamma radiation is penetrating and can affect most electrical equipment. Simple equipment (like motors, switches, incandescent lights, wiring, and solenoids) is very radiation resistant and may never show any radiation effects, even after a very large radiation exposure. Diodes and computer chips (electronics) are much more sensitive to gamma radiation. To give you a comparison of effects, it takes a radiation dose of about 5 Sv to cause death to most people. Diodes and computer chips will show very little functional detriment up to about 50 to 100 Sv. Also, some electronics can be "hardened" (made to be not affected as much by larger gamma radiation doses) by providing shielding or by selecting radiation-resistant materials.

 

Some electronics do exhibit a recovery after being exposed to gamma radiation, after the radiation is stopped. But the recovery is hardly ever back to 100% functionality. Also, if the electronics are exposed to gamma radiation while unpowered, the gamma radiation effects are less.

 

Ionizing radiation breaks down the materials within the electrical equipment. For example, when wiring is exposed to gamma rays, no change is noticed until the wiring is flexed or bent. The wire's insulation becomes brittle and will break and may cause shorts in the equipment. The effect on diodes and computer chips is a bit more complex. The gamma rays disrupt the crystalline nature of the inside of the electronic component. Its function is degraded and then fails as more gamma radiation exposure is received by the electronic component.

 

Gamma rays do not affect the signals within the device or the signals received by the device. Nonionizing radiation (like radio signals, microwaves, and electromagnetic pulses) DO mess with the signals within and received by the device. I put a cheap electronic game in my microwave oven at home. It arced and sparked and was totally ruined. I didn’t waste any more of my time playing that game.

Hope this helps.

Explanation:

MARK ME AS BARINIEST PLS

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You perform a distillation to separate a mixture of propylbenzene and cyclohexane, and you obtain 2.9949 grams of cyclohexane (d
Romashka-Z-Leto [24]

Answer:

66.67%

Explanation:

From the given information:

mass of cyclohexane = 2.9949 grams

density of cyclohexane = 0.779 g/mL

Recall that:

Density = mass/volume

∴

Volume = mass/density

So, the volume of cyclohexane = 2.9949 g/ 0.779 g/mL

= 3.8445 mL

Also,

mass of propylbenzene = 1.6575 grams

density of propylbenzene = 0.862 g/mL

Volume of propylbenzene =  1.6575 g/ 0.862 g/mL

= 1.9229 mL

The volume % composition of cyclohexane from the mixture is:

= (\dfrac{v_{cyclohexane}}{v_{cyclohexane}+v_{propylbenzene}})\times 100

= (\dfrac{3.8445}{3.8445+1.9229})\times 100

= (\dfrac{3.8445}{5.7674})\times 100

= 66.67%

6 0
3 years ago
38 points!! >:D
Sav [38]

Answer: which*

Explanation:

8 0
3 years ago
Read 2 more answers
How many joules of heat must be absorbed by 500g h2O @ 50CELCIUS to convert to steam @ 120 celcius?
nadya68 [22]

Answer:

Q = 1267720 J

Explanation:

  • Qt = QH2O + ΔHv

∴ QH2O = mCpΔT

∴ m H2O = 500 g

∴ Cp H2O = 4.186 J/g°C = 4.183 E-3 KJ/g°C

∴ ΔT = 120 - 50 = 70°C

⇒ QH2O = (500 g)(4.183 E-3 KJ/g°C)(70°C) = 146.51 KJ

∴ ΔHv H2O = 40.7 KJ/mol

moles H2O:

∴ mm H2O = 18.015 g/mol

⇒ moles H2O = (500 g)(mol/18.015 g) = 27.548 mol H2O

⇒ ΔHv H2O = (40.7 KJ/mol)(27.548 mol) = 1121.21 KJ

⇒ Qt = 146.51 KJ + 1121.21 KJ = 1267.72 KJ = 1267720 J

5 0
3 years ago
Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m
erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3}  =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2}  =0.06882molBr^-

Now, we compute the total moles of bromide:

n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol

Then, the total volume in liters:

150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L

Therefore, the concentration of total bromide is:

M=\frac{0.12338mol}{0.325L}\\\\M=0.380M

Best regards!

8 0
2 years ago
How many moles of h2o will be produced from 2.9 g of HCl reacting with Ca(OH)2?
Igoryamba

Answer:

0.080 mol

Explanation:

 M(HCl) = (1.0 +35.5) g/mol = 36.5 g/mol

2.9g*1mol/36.5 g = 0.0795 mol HCl

                              Ca(OH)2 + 2HCl ---> CaCl2 + 2H2O

from reaction                            2 mol                   2 mol

given                                       0.0795 mol           x mol

x = 0.0795 mol ≈0.080 mol

7 0
3 years ago
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