P27_009.PDF

9. (a) The charge that strikes the surface in time ∆ t is given by ∆ q = i ∆ t ,where i is the current. Since

each particle carries charge 2 e , the number of particles that strike the surface is

N = ∆ q

2 e = i ∆ t

= (0 . 25

10 − 6 A)(3 . 0s)

=2 . 3

10 12

2 e

2(1 . 6

10 − 19 C)

(b) Now let N be the number of particles in a length L of the beam. They will all pass through the

beam cross section at one end in time t = L/v ,where v is the particle speed. The current is the

charge that moves through the cross section per unit time. That is, i =2 eN/t =2 eNv/L .Thus

N = iL/ 2 ev . To ﬁnd the particle speed, we note the kinetic energy of a particle is

10 − 12 J .

Since K = 2 mv 2 , then the speed is v = 2 K/m . The mass of an alpha particle is (very nearly) 4

times the mass of a proton, or m =4(1 . 67

K =20MeV=(20

10 6 eV)(1 . 60

10 − 19 J / eV) = 3 . 2

10 − 27 kg) = 6 . 68

10 − 27 kg, so

v =

2(3 . 2

10 − 27 kg =3 . 1

10 − 12 J)

10 7 m / s

6 . 68

and

2 ev =

(0 . 25

10 − 6 )(20

10 − 2 m)

N =

10 7 m / s) =5 . 0

10 3 .

2(1 . 60

10 − 19 C)(3 . 1

10 − 12 J. We note, too, that the initial potential energy is U i = qV =2 eV ,and

the ﬁnal potential energy is zero. Here V is the electric potential through which the particles are

accelerated. Consequently,

K f = U i =2 eV =

⇒

V = K f

2 e =

3 . 2

10 − 12 J

10 − 19 C) =10

10 6 V .

2(1 . 60

is 20MeV = 3 . 2

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