hw4 - CHAPTER3 ATOMS 39 Drrc(200 Q0pn xy'to = liz pnl 4...

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CHAPTER3. ATOMS 39 3.38 r /newJ Other things bei"g_gqgrl D, , n tft. Therefore, if 4,.s(20 s) : r0 p.m, then Drrc(200 4 : Q0 pn) xy'to = liz pnl 3.39 ro [new] At given temperature and viscosity, Eq. (3.36) implies that D x 1/tla, (where D is the rms displacement) whence t u. D2a. Thus, if a2:100ar and Dz:100Dt,thent2:106xt1 = 2 x 10? s = Fioltnr I. That is, it would take a particle that is visible to the naked eye about 8 months to move a clearly visible distance, 3.40 rr [old 4.19] Time, t: tlu, = l7"l ttl* as = -eE/m, we find u, : ant:F"Ell;4 Finally, d = arctan(-uv /u,) = -uu/u-, [email protected] 3.41 or [old 4.20] The magnetic field bends the electrons in a circular path, as shown, with radius given by Eq. (2.48) w R : muleB , The deflection angle 0 = s/R: seB/(mu), or, since s = I because d is small, F;[email protected] 3.42 oo [new] The dependence on o is surprisingly complicated: In the electric case, the time t spent in the field is inversely proportional to the initial speed o. The acquired sideways velocrry uu b proportional to I and hence also inversely proportional to o; that is, uu contains a factor of s in the denominator. The angular deflection d is (approximat ely) unfu, which gives it a second factor of o in the denominator. In the electric case, this is the end of the story and d x 1fu2, ln the magnetic case, the same considerations apply except that, in addition, the force is proportional to o, and this puts a factor of u in the numerator of u, canceling one in the denominatorl thus, finallg e u.7/u. 3.4$ or [old 4.21] Let.EI be the heat released, f/ = (number of electrons) x (KE of each electron) : (Q/d " @u2 121, 6 fi = Qmu2l(2e). Flom Problem 3.41, 0 x teBl(mu), and multiplying these last two equations together, we find Ild x QulB/2, whence lr., =zHll(QIB)1. we ca"n now solve the result of problem - 3.1r fot mf e = B/(0u) xletr82/(20rH)l t44ro [old 4.22] The two equations aremu2f2: evo and. n: mu/(eB). we can solve the second mle : R2 B2 /(zVs) frr o, to give o : ReB
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40 CHAPTER3. ATOMS 3.45 ccc {old 4.231 (a) Adding the two equations (3.67), we find r-ra * tr,, : qE/(6nrd, and putting u: t /t, 11oE y:-q-+i=,tra (I) where we have introduced g to denote the combination (1lti + Qlt"). (b) Using the average of the ten values of rd, we get td =
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hw4 - CHAPTER3 ATOMS 39 Drrc(200 Q0pn xy'to = liz pnl 4...

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