If the total energy transferred to a surface in time $$t$$ is $$6.48 \times 10^5$$ J, then the magnitude of the total momentum delivered to this surface for complete absorption will be:

To find the magnitude of the total momentum delivered to the surface for complete absorption, we use the relationship between the energy of electromagnetic radiation and its momentum.

When radiation is completely absorbed by a surface, the total momentum $$p$$ delivered to it is given by the formula:

$$p = \frac{E}{c}$$

where:

• $$E$$ is the total energy transferred to the surface $$= 6.48 \times 10^5\text{ J}$$
• $$c$$ is the speed of light in vacuum $$\approx 3 \times 10^8\text{ m/s}$$

Substituting the given values into the formula:

$$p = \frac{6.48 \times 10^5}{3 \times 10^8}$$

$$p = 2.16 \times 10^{-3}\text{ kg m s}^{-1}$$