Given, ${R}_1=\left\{\left(p, p^n\right): p\right.$ is a Prime and $n \geq 0$ is an integer $\}$

and, set $A=\{1,2,3 \ldots \ldots .50\}$

$p$ is a Prime number which can take 15 values $2,3,5,7,11,13,17,19,23,29,31,37,41,43$ and 47

$\therefore$ We can calculate no. of elements in $\mathrm{R}_1$

$$ \mathrm{R}_1=\left(2,2^0\right),\left(2,2^1\right),\left(2,2^2\right)\left(2,2^3\right) \ldots \ldots\left(2,2^5\right)=6$$ number of ordered pairs

$\left(3,3^0\right),\left(3,3^1\right),\left(3,3^2\right) \ldots \ldots . .\left(3,3^3\right)=4$ number of order paris

$\left(5,5^0\right),\left(5,5^1\right),\left(5,5^2\right) \ldots \ldots \ldots . .=3$ number of order paris

$\left(7,7^0\right) \ldots \ldots .\left(7,7^2\right) \ldots \ldots \ldots=3$ number of order paris

$\left(11,11^0\right)$ and $\left(11,11^1\right)=2$ number of order paris

$\left(13,13^0\right)$ and $\left(13,13^1\right)=2$ number of order paris

$$ \therefore $$ For the 11 prime numbers ($11,13,17,19,23,29,31,37,41,43$ and 47), $n$ can only be 0, 1 (two pairs each).

$$ \therefore n\left(\mathrm{R}_1\right)=6+4+3+3+(2 \times 11)=38 $$

$\mathrm{R}_2=\left(p, p^n\right) $, where n = 0 or 1

$$ \left(2,2^0\right),\left(2,2^1\right)\left(3,3^0\right)\left(3,3^1\right) \ldots . .\left(47,47^0\right)\left(47,47^1\right) $$

Two ordered pairs of each element $n\left({R}_2\right)=2 \times 15=30$ elements

Hence $ R_1-R_2=38-30=8$