Question

Alpha particles which have a charge +2e and mass 6.64 ✕ 10−27 kg are initially at rest and are fired directly at a stationary lead nucleus (charge +82e) at a speed of 1.95 ✕ 107 m/s. Determine the distance of closest approach (in fm) before the alpha particle reverses direction. Assume the lead nucleus remains stationary. Assume the alpha particles are initially very far from a stationary lead nucleus.

Answer 1

Answer:

The closest approach distance is 29.9 fm

Solution:

As per the question:

Charge on alpha-particle, q = +2e C

Charge on Lead nucleus, q' = +82e C

Speed of the $\alpha-particle$, $v_{\alpha} = 1.95\times 10^{7} m/s$

Mass of $\alpha-particle$, M = $6.64\times 10^{- 27} kg$

Now,

The kinetic energy of the $\alpha-particle$:

$KE = \frac{1}{2}mv_{\alpha^{2}}$

Also the electro-static potential energy of the $\alpha-particle$ is given by:

$U_{E} = \frac{1}{4\piepsilon_{o}}\frac{qq'}{D}$

where

D = distance between the nucleus of $\alpha-particle$ and lead.

Since, the kinetic energy of the $\alpha-particle$ get tranformed into electro-static potential energy, thus equating the two:

$\frac{1}{2}mv_{\alpha^{2}} = \frac{1}{4\piepsilon_{o}}\frac{qq'}{D}$

$\frac{1}{2}\times 6.64\times 10^{- 27}\times (1.95\times 10^{7})^{2} = (9\times 10^(9))\frac{2e.82e}{D}$

$D = (9\times 10^(9))\frac{164\times (1.6\times 10^{- 19})^{2}}{1.26\times 10^{- 12}} = 2.99\times 10^{- 14} m = 2.99\times 10^{- 14} m = 29.9 fm$

Answer 2

Alpha particle is positive charged radioactive particle. The distance of closest approach (in fm) before the alpha particle reverses direction is 29.9 fm.

What is alpha particles?

Alpha particles is the particle in which the number of protons is 2 and the number of neutron is 2. These protons and neutron in alpha particle bound together tightly.

Given information-

The charge if alpha particles is +2e.

The mass of the alpha particles is $6.64\times10^{-27}$ kg.

The initial speed of the particles is $1.95\times10^{7}\rm m/s$.

As the kinetic energy of the alpha particles is equal to its the potential energy. Thus,

$\dfrac{1}{2}mv^2=k\dfrac{qq^'}{D}$

Put the values as,

$\dfrac{1}{2}\times6.64\times10^{-27}\times(1.95\times10^7)^2=9\times10^9\dfrac{2e\times 82e}{D}\\D=29.9\rm fm$

Thus the distance of closest approach (in fm) before the alpha particle reverses direction is 29.9 fm.

Learn more about the alpha particle here;

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