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torisob [31]
3 years ago
10

Find the minimum kinetic energy in MeV necessary for an α particle to touch a 39 19 K nucleus that is initially at rest, assumin

g only Coulomb forces are present. Assume that the α particle has a radius of 2.0 fm, and the potassium nucleus of 3.7 fm, and take into account the masses of both particles.

Physics
1 answer:
SIZIF [17.4K]3 years ago
4 0

Answer:

KE = 9.6MeV

Explanation:

Given the relationships we understand that,

q_1 = 19e, q_2 = 2e

The Potential at point p, is given by the following formula

V_ {p} = \frac {Kq_1} {r}

According to the graphic designed, you have,

V_ {p} = \frac {Kq_1} {(r_1 + r_2)}

V_ {p} = \frac {(9 * 10 ^ 9) (19 * 1.6 * 10 ^ {- 15})} {(3.7 + 2) * 10 ^{- 15}}}

V_ {p} = 4.8 * 10 ^ 6V

The kinetic energy of the particle would be given by

KE = q_2 * V_ {p} = 2e * 4.8 * 10 ^ 6V

KE = 9.6MeV

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An electromagnetic wave of wavelength 435 nm is traveling in vacuum in the —z direction. The electric field has an amplitude of
Aloiza [94]

Answer:

a) 6.9*10^14 Hz

b) 9*10^-12 T

Explanation:

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a)

To find the frequency of the wave, we would be applying this formula

c = fλ, where c = speed of light

f = c/λ

f = 3*10^8 / 435*10^-9

f = 6.90*10^14 Hz

b) again, to find the amplitude of the magnetic field, we would use this relation

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B(max) = 2.7*10^-3 / 3*10^8

B(max) = 9*10^-12 T

c) and lastly,

1T = 1 (V.s/m^2)

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3 years ago
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