Question

Considering that the alpha-particles carry average kinetic energy of 2.00 x 1010 J,

calculate the maximum size of the gold nucleus. [Atomic number of gold is 79 and

e= 1.60 x 10-19C]​

Answer 1

Explanation:

number of basic units of charge, e. ... each droplet are all integer multiples of -1.60χ10-19 C. We therefore state ... Atomic and Nuclear Physics – Page 2 of 14 ... Answer is 8.5 × 1010 J.

Answer 2

Answer:

The radius of gold nucleus is $r=9.1\times10^{-39}m$

Explanation:

Given that,

Alpha particles carry average kinetic energy of $2\times10^{10}J$ and the atomic number of gold is $Z=79$ .

We are required to find the maximum possible radius of gold nucleus for which we shall use the kinetic energy relation for an α particle as shown below-

$K.E=\frac{1}{4\pi \epsilon_{0}} \frac{Ze^{2}}{r}$

Substituting the values of kinetic energy,atomic number and charge in above relation,we get-

$2\times10^{10}=9\times10^{9}\times\frac{79(1.6\times10^{-19})^{2}}{r} \\= > 20=\frac{711\times2.56\times10^{-38}}{r} \\= > r=9.1\times10^{-39}m$

Therefore,the radius of gold nucleus is $r=9.1\times10^{-39}m$

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