Question

1
In scattering experiment, find the distance of closest approach, if a 6 MeV a-particle is used
(1) 3.2 * 10-16 m
(2) 2 x 10-14 m
(3) 4.6 x 10-15 m
(4) 3.2 * 10-15 m​

Answer 1

Answer:

Closest distance of approach of alpha particle is

$r = 3.8 \times 10^{-14} m$

Explanation:

As we know that during scattering experiment the kinetic energy of alpha particle will convert into the electrostatic potential energy of the charge system of alpha particle and gold nuclei when it reaches to closest distance

So here we have

$KE = \frac{kq_1q_2}{r}$

so we have

$6 \times 10^6 (1.6 \times 10^{-19}) = \frac{(9\times 10^9)(2\times 1.6 \times 10^{-19})(79 \times 1.6 \times 10^{-19})}{r}$

so we have

$r = 3.8 \times 10^{-14} m$

#Learn

Topic : Electrostatic potential energy

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