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- The electric potential difference AV, also called voltage, is expressed in terms of the electric field as follows AV = - Edr a where, a is the initial position and b is the final position. If an electric field in a region in space is defined as E= - Br2 where B is just a constant without an arbitrary value. What is the electric potential difference from position 0 to R? 1 AV = - BR 1 AV = 2 AV = Br O AV = - BR³ O AV = -(1- B)aR³HW What is the Linear Energy Transfer resulting from the passage of a 0.1-MeV beta particle through standard air (where NZ=3.88x102º electron/cm² and I=8.6x10$ for air) use the relativity for the kinetic energy of electronThe electric potential difference AV, also called voltage, is expressed in terms of the electric field as follows AV = | Edr a where, a is the initial position and b is the final position. If an electric field in a region in space is defined as E= Br where B is just a constant without an arbitrary value. What is the electric potential difference from position 0 to R? 1 AV = BR3 3 1 AV = BR3 AV = 1 (1– B)aR³ 2 AV = Br3 O AV = -(1-B)aR³ 1 AV = Br3 1 AV = BR2
- hd Px 5. Evaluate the following commutators: (Remember that i dx ). (a) [x.y] (b) [x.px] (c) [ps.py] (d) [x².px] (e) [(1/x).px] (f) [(1/x).p²] (g) [xpy-ypx.yp-zpyGiven: Nc = (2.51x1019)(mn/mo)3/2 (T/300)3/2 Nv = (2.51x1019)(mp/mo)3/2 (T/300)3/2 and ni = (NcNv)1/2 e(-Eg/2kT) show that: ni = (2.51x1019) ((mn/mo) * (mp/mo) )3/2 e(-Eg/2kT)decay in the discharging process to half of its maximum (V = Vo/2). This time is the half-life: t1= t1/2. 20.82 ms, 3.977v T1= 1.4 m/s Measure the time it takes for the voltage to decay to one-quarter (V = Vo/4 = (Vo/2)2) of its maximum. This is two half-lives: t2 = 2t1/2. Then divide this time by two to find the half-life (t1/2 = t2/2) 20.2-24.00 = 3.2 T2 = 3.2/2 = 1.6m/s Measure the time it takes for the voltage to grow to one-eighth (V = Vo/8 = (Vo/2)3) of its This is three half-lives, t3 = 3t1/2. Then divide this time by three to find the half-life, t1/2 = t3/3. 20.8 – 22.28= 1.48m/s T3= 1.48/3 = 0.49 m/s Take the average of the three measured values of the half-life. Estimate the precision of the measurement and state it as {half-life ± precision}…
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- 12QM. Please answer question throughly and detailed.Suppose the range for 5.0 MeVα ray is known to be 2.0 mm in a certain material. Does this mean that every 5.0 MeVα 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.MV². -)dV 2RT M dN = 47(- N 3/2v² exp(- 2 RT F(V)dV =! Find the Vm value by considering the following curve 25 x 10-* 20 x 10-4 298 K 15 x 10-4 10 x 10-4 1500 K 5 x 10-4 Vm Vm u, m s-1 All questions are compulsory. 00s 000z 00S np N NP I