InterviewSolution
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InterviewSolution
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A steady beam of `alpha`-particles travelling with kinetic energy `E=83.5 keV` carries a current of `I=0.2 mu A`. Mass of `alpha`-particle `=6.68 xx10^(-27)kg`. (i) If this beam strikes a plane surface at an angle `theta=60^@` with normal to the surface, how many `alpha` -particles strike the surface in t=4 second? |
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Answer» Here, current `I = 0.25 mu A = 0.25 xx 10^(-6)A` Kinetic energy of `alpha`-particle, `K = 84 KeV = 84 xx 10^(3) xx 1.6 xx 10^(-19) J` `= 84 xx 1.6 xx 10^(-16)J` (a) Charge on `alpha`-particles striking the plane surface per second `n = I/q = ( 0.25 xx 10^(-6))/( 2 xx 1.6 xx 10^(-19)) = 7.81 xx 10^(11) s^(-1)` Number of `alpha`-particles striking the surcface in `4 s = nt = (7. 81 xx 10^(11)) xx 4` `= 31.24 xx 10^(11) = 3.1 xx 10^(12)` (b) Let v be the velocity of `alpha`-particle when they are travelling towards the surface. Then `K = 1/2 mv^(2)` or `v = sqrt((2 K)/(m)) = sqrt((2 xx 84 xx 1.6 xx10^(-16))/(6.68 xx 10^(-27))) = 2 xx 10^(6) ms^(-1)` It shows that a beam of length `2 xx 10^(6)` m crosses a section in one second. But numbere of `alpha`-particle passing through the section in one second is `n = 7.81 xx 10^(11) s^(-1)`. Therefore, no. of `alpha`-particles in unit length of beam ` =n/v` Hence no. `alpha`-particles in length l of the beam `= (nl)/(v) = ((7. 81 xx 10^(11)) xx 0.30)/(2 xx 10^(6)) = 1.17 xx 10^(5)` |
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