$$ Q = n\,C_P\,\Delta T \quad \text{and} \quad \Delta U = n\,C_V\,\Delta T $$

For a monoatomic ideal gas, the molar specific heats are given by:

$$ C_V = \frac{3}{2}R \quad \text{and} \quad C_P = C_V + R = \frac{5}{2}R $$

Thus, the ratio of the heat added to the change in internal energy is:

$$ \frac{Q}{\Delta U} = \frac{C_P}{C_V} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3} $$

Given that:

$$ \frac{E_1}{E_2} = \frac{x}{9} $$

We equate the two ratios:

$$ \frac{5}{3} = \frac{x}{9} $$

Multiplying both sides by 9:

$$ x = \frac{5}{3} \times 9 = 15 $$

Thus, the value of $$x$$ is 15.