The radius of a nucleus of mass number 64 is 4.8 fermi. Then the mass number of another nucleus having radius of 4 fermi is $\frac{1000}{x}$, where $x$ is _______.

<p>According to the empirical formula relating the radius of a nucleus ($ R $) with its mass number ($ A $), we know that the radius of a nucleus is proportional to the cube root of its mass number. This relationship is given as:</p> <p>$R = R_0 A^{1/3}$</p> <p>where $ R_0 $ is a constant with an approximate value of 1.2 fermis.</p> <p>Given that for a nucleus with a mass number 64 has a radius of 4.8 fermis, we can write:</p> <p>$4.8 \text{ fermi} = R_0 \times 64^{1/3}$</p> <p>Now another nucleus has a radius of 4 fermis:</p> <p>$4 \text{ fermi} = R_0 \times A'^{1/3}$</p> <p>Where $ A' $ is the mass number of the other nucleus.</p> <p>Let's solve for $ R_0 $ from the first equation:</p> <p>$R_0 = \frac{4.8 \text{ fermi}}{64^{1/3}}$</p> <p>Now we're going to find the mass number $ A' $ using the second equation and substituting $ R_0 $ from the above:</p> <p>$$ 4 \text{ fermi} = \left( \frac{4.8 \text{ fermi}}{64^{1/3}} \right) \times A'^{1/3} $$</p> <p>Now, we want to find $ A' $ in terms of $ x $ as given by the equation in the question:</p> <p>$A' = \frac{1000}{x}$</p> <p>Substitute $ A' $ in the equation above, we get:</p> <p>$$ 4 = \left( \frac{4.8}{64^{1/3}} \right) \times \left( \frac{1000}{x} \right)^{1/3} $$</p> <p>Let's solve for $ x $:</p> <p>$(4)^3 = \left( \frac{4.8}{64^{1/3}} \right)^3 \times \frac{1000}{x}$</p> <p>$4^3 \times x = \left( \frac{4.8}{64^{1/3}} \right)^3 \times 1000$</p> <p>$x = \left( \frac{\left( \frac{4.8}{64^{1/3}} \right)^3 \times 1000}{4^3} \right)$</p> <p>Now calculate the values:</p> <p>$x = \left( \frac{\left( \frac{4.8}{4} \right)^3 \times 1000}{64} \right)$</p> <p>$x = \left( \frac{1.2^3 \times 1000}{64} \right)$</p> <p>$x = \left( \frac{1.728 \times 1000}{64} \right)$</p> <p>$x = \left( \frac{1728}{64} \right)$</p> <p>$x = 27$</p> <p>Therefore, the value of $ x $ is 27.</p>