To calculate the kinetic energy ($K.E.$) of the hydrogen ion, we use the formula:

$K.E. = \frac{1}{2} m v^2$

where:

• $m$ is the mass of the hydrogen ion ($1.6726219 \times 10^{-27}$ kg),
• $v$ is the velocity of the hydrogen ion ($8.175 \times 10^7$ m/s).

Now, substituting the given values into the formula:

$K.E. = \frac{1}{2} \times 1.6726219 \times 10^{-27} \, \text{kg} \times (8.175 \times 10^7 \, \text{m/s})^2$

First, calculate the square of the velocity:

$v^2 = (8.175 \times 10^7)^2 = 66.82203125 \times 10^{14} \, \text{m}^2/\text{s}^2$

Then, multiply by the mass and divide by 2:

$K.E. = \frac{1}{2} \times 1.6726219 \times 10^{-27} \times 66.82203125 \times 10^{14}$

Simplifying further:

$K.E. = \frac{1}{2} \times 111.75706577 \times 10^{-13} \, \text{J}$ $K.E. = 55.878532885 \times 10^{-13} \, \text{J}$

Finally, expressing the result with four significant figures and in scientific notation:

$K.E. \approx 5.588 \times 10^{-12} \, \text{J}$

So, the kinetic energy of the hydrogen ion is approximately $5.588 \times 10^{-12}$ joules.

Would you like further details or have any questions? Here are five related questions that might interest you:

1. How does the kinetic energy of particles in the solar wind compare to those in Earth's atmosphere?
2. How does the velocity of a particle affect its kinetic energy?
3. What role does kinetic energy play in the solar wind's interaction with planetary atmospheres?
4. How is the mass loss of the Sun due to solar wind calculated?
5. How does the speed of solar wind particles vary across different regions of the Sun?

Tip: Always remember that in physics, energy is often expressed in scientific notation to handle the extremely small or large quantities commonly encountered.