Suppose the range for 5.0 MeVa ray is known to be 2.0 mm in a certain material. Does this mean that every 5.0 MeVa a ray that strikes this material travels 2.0 mm, or does the range have an average value with some statistical fluctuations in the distances traveled? Explain.
Suppose the range for 5.0 MeVa ray is known to be 2.0 mm in a certain material. Does this mean that every 5.0 MeVa a ray that strikes this material travels 2.0 mm, or does the range have an average value with some statistical fluctuations in the distances traveled? Explain.
Suppose the range for 5.0 MeVa ray is known to be 2.0 mm in a certain material. Does this mean that every 5.0 MeVa a ray that strikes this material travels 2.0 mm, or does the range have an average value with some statistical fluctuations in the distances traveled? Explain.
Expert Solution & Answer
To determine
Whether it means that every 5.0 MeVαray that strikes this material travels 2.0mm , or the range have an average value with some statistical fluctuations in the distances traveled, suppose the range for 5.0 MeVα ray is known to be 2.0mm in a certain material
Answer to Problem 1CQ
Every 5.0 MeVa a ray that strikes this material won't travel 2.0 mm.
Explanation of Solution
Concept Used:
Nuclear radioactivity.
The distance travel by the radiation through a material is defined as the range of the radiation. The range of radiation depends upon some of the factors which includes the energy of the radiation, the material through which it travels and the type of the radiation whether alpha, beta or gamma ray. By defining these factors or by knowing these factors we can know the range of the radiation. Here it is given that range of radiation is 2 mm for a certain material, But the charged particles in the material interacts with rays and because of that ray will show some random fluctuations, so every 5.0 MeVαray won't travel 2.0mm on the same material. In the α decay the energy released in it is about MeV range. It is about 106 for a typical chemical reaction. Most of the energy is converted in kinetic energy of the α particle which moves at high speed
Conclusion:
Thus, every 5.0 MeVa a ray that strikes this material won't travel 2.0 mm.
Want to see more full solutions like this?
Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
5. An XPS with an instrument work function of 4.5eV was used to analyze a surface. An observed kinetic energy of 795.2 eV was observed when a(n) MgKα (1253.6 eV) X-ray source was used. What was the binding energy of the electron coming from this surface?
VBE = 0.7V, β=hfe = 100, determine Av = Vo/Vi
- reg YEARY PC PB YD TEMP PRP
Source
df
MS
Number of obs
29
F(6, 22)
=
1777.52
Model
2025.82114
337.636856
Prob > F
0.0000
=
Residual
R-squared
Adj R-squared
4.17886221
22
.189948282
0.9979
0.9974
Total
2030
28
72.5
Root MSE
=
.43583
YEAR
Coefficient
Std. err.
t
P>|t|
[95% conf. interval]
Y
.0640617
.0523215
1.22
0.234
-.0444465
.1725699
PC
-.0179212
.010265
-1.75
0.095
-.0392094
.003367
PB
.0058838
.0050092
1.17
0.253
-.0045047
.0162722
YD
.0978191
.0142261
6.88
0.000
.068316
.1273222
TEMP
.0021932
.0054143
0.41
0.689
-.0090354
.0134218
PRP
.0087202
.0091906
0.95
0.353
-.01034
.0277804
cons
1966.475
1.727965
1138.03
0.000
1962.892
1970.059
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.