Vertical beam of mercury (Hg atomic mass =202u) atoms emerges from the oven kept at T= 400K. Consider three atoms, emerging from the oven: one with the average speed, one with the rms speed, and the other with the most probable speed of the Maxwellian distribution
a) Find the speeds of each atom as they emerge from the oven
b) Find the maximum vertical position reached the fastest atom
c) Find the distance between highest positions fastest of the three and the slowest of the three atoms
To answer these questions, we need to understand the Maxwellian distribution and the behavior of atoms at a given temperature. The Maxwellian distribution describes the speed distribution of particles in a gas at a given temperature. It provides the probabilities of particles having different speeds.
a) Finding the speeds of each atom as they emerge from the oven:
To determine the speeds, we will use the Maxwellian distribution. The most probable speed (Vmp), average speed (Vavg), and root mean square speed (Vrms) can be calculated using the following formulas:
Vmp = (2kT / m)^(1/2)
Vavg = (8kT / πm)^(1/2)
Vrms = (3kT / m)^(1/2)
Where:
- k is the Boltzmann constant (k = 1.38 x 10^-23 J/K)
- T is the temperature in Kelvin (T = 400K)
- m is the mass of a mercury atom (m = 202u = 202 x 1.66 x 10^-27 kg)
Plugging in the given values into the formulas, we can calculate the speeds:
Vmp = (2 * 1.38 x 10^-23 J/K * 400K / (202 x 1.66 x 10^-27 kg))^(1/2)
Vavg = (8 * 1.38 x 10^-23 J/K * 400K / π * (202 x 1.66 x 10^-27 kg))^(1/2)
Vrms = (3 * 1.38 x 10^-23 J/K * 400K / (202 x 1.66 x 10^-27 kg))^(1/2)
Calculating these values, we can find the speeds of each atom as they emerge from the oven.
b) Finding the maximum vertical position reached by the fastest atom:
To determine the maximum vertical position reached, we can make use of the equations of motion. We need to consider the effects of gravity and the initial speed of the fastest atom (Vmp). Assuming the atoms move in a vacuum and neglecting air resistance, we can use the following equation:
h = (V0^2 / (2g))
Where:
- h is the maximum vertical position reached
- V0 is the initial vertical velocity of the fastest atom (Vmp)
- g is the acceleration due to gravity (g = 9.8 m/s^2)
Plugging in the values, we can calculate the maximum vertical position reached by the fastest atom.
c) Finding the distance between the highest positions of the fastest and slowest atoms:
To find the distance between the highest positions of the fastest and slowest atoms, we need to consider the difference in their initial speeds (Vmp and Vavg). We can again use the equations of motion to find the difference in vertical positions. Assuming both atoms start from the same initial height and neglecting air resistance, we can calculate the difference using the equation:
Δh = (V1^2 - V2^2) / (2g)
Where:
- Δh is the difference in vertical positions
- V1 and V2 are the initial speeds of the fastest (Vmp) and slowest (Vavg) atoms, respectively
- g is the acceleration due to gravity (g = 9.8 m/s^2)
Plugging in the values, we can calculate the difference in vertical positions between the fastest and slowest atoms.
By following these steps and using the formulas provided, you can find the answers to the questions regarding the speeds, maximum vertical position, and the distance between the highest positions of the three atoms.