QJEEM24Phf63f: JEE Main 2024 Physics - Class 12 JEEM24Phf63f Solution: The radius of a nucleus of mass number 64 is 4.8 fermi. Then…

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:

$R = R_0 A^{1/3}$

where $R_0$ is a constant with an approximate value of 1.2 fermis.

Given that for a nucleus with a mass number 64 has a radius of 4.8 fermis, we can write:

$4.8 \text{ fermi} = R_0 \times 64^{1/3}$

Now another nucleus has a radius of 4 fermis:

$4 \text{ fermi} = R_0 \times A'^{1/3}$

Where $A'$ is the mass number of the other nucleus.

Let's solve for $R_0$ from the first equation:

$R_0 = \frac{4.8 \text{ fermi}}{64^{1/3}}$

Now we're going to find the mass number $A'$ using the second equation and substituting $R_0$ from the above:

$4 \text{ fermi} = \left( \frac{4.8 \text{ fermi}}{64^{1/3}} \right) \times A'^{1/3}$

Now, we want to find $A'$ in terms of $x$ as given by the equation in the question:

$A' = \frac{1000}{x}$

Substitute $A'$ in the equation above, we get:

$4 = \left( \frac{4.8}{64^{1/3}} \right) \times \left( \frac{1000}{x} \right)^{1/3}$

Let's solve for $x$ :

$(4)^3 = \left( \frac{4.8}{64^{1/3}} \right)^3 \times \frac{1000}{x}$

$4^3 \times x = \left( \frac{4.8}{64^{1/3}} \right)^3 \times 1000$

$x = \left( \frac{\left( \frac{4.8}{64^{1/3}} \right)^3 \times 1000}{4^3} \right)$

Now calculate the values:

$x = \left( \frac{\left( \frac{4.8}{4} \right)^3 \times 1000}{64} \right)$

$x = \left( \frac{1.2^3 \times 1000}{64} \right)$

$x = \left( \frac{1.728 \times 1000}{64} \right)$

$x = \left( \frac{1728}{64} \right)$

$x = 27$

Therefore, the value of $x$ is 27.