In the Bohr model of the hydrogen atom, an electon mass= 9.11 x 10^-31 kg orbits a proton at a radius of 5.29 x 10^-11 m. The proton pulls on the electron with a force of 8.20 x 10^-8N. what is the speed of the electron? what is the period of the electron?
I really need help setting this up.
F = m v^2/r
does this look right...
mg + force of proton = mv^2/r
9.8m/s2 + 8.20 x 10^-8n = v^2/r
8.20 x 10^-8 (5.29 x 10^-11) = v^2
4.34 x 10^-18 = v^2
v= 2.08 x 10^-9
this is for the speed
To find the speed of the electron in the Bohr model of the hydrogen atom, you can use the centripetal force equation:
F = (m * v^2) / r
Where:
F = force (given as 8.20 x 10^-8 N)
m = mass of the electron (given as 9.11 x 10^-31 kg)
v = speed of the electron (unknown)
r = radius of the orbit (given as 5.29 x 10^-11 m)
Now let's solve for the speed of the electron:
1. Rearrange the equation to solve for v:
v = sqrt((F * r) / m)
2. Substitute the given values into the equation:
v = sqrt((8.20 x 10^-8 N * 5.29 x 10^-11 m) / (9.11 x 10^-31 kg))
3. Simplify the equation:
v = sqrt(3.9268 x 10^-18 m^2 / kg)
4. Calculate the square root:
v ≈ 6.276 x 10^5 m/s
Therefore, the speed of the electron in the Bohr model of the hydrogen atom is approximately 6.276 x 10^5 m/s.
Now, let's calculate the period of the electron using the formula:
T = (2πr) / v
Where:
T = period of the electron (unknown)
r = radius of the orbit (given as 5.29 x 10^-11 m)
v = speed of the electron (calculated as 6.276 x 10^5 m/s)
Let's find the period of the electron:
1. Substitute the given values into the equation:
T = (2π * 5.29 x 10^-11 m) / (6.276 x 10^5 m/s)
2. Simplify the equation:
T = (3.3237 x 10^-10 m) / (6.276 x 10^5 m/s)
3. Divide the values:
T ≈ 5.301 x 10^-16 s
Therefore, the period of the electron in the Bohr model of the hydrogen atom is approximately 5.301 x 10^-16 s.
To determine the speed of the electron and the period of its orbit in the Bohr model, we can start by recognizing that the electron is moving in a circular orbit around the proton. The force between them is provided, so we can use the centripetal force formula to solve for the speed and period of the electron.
The formula for centripetal force is:
F = (m * v^2) / r
Where:
F = force acting on the electron (8.20 x 10^-8 N)
m = mass of the electron (9.11 x 10^-31 kg)
v = speed of the electron (what we want to find)
r = radius of the electron's orbit (5.29 x 10^-11 m)
To solve for the speed of the electron, we can rearrange the formula:
v^2 = (F * r) / m
v = √((F * r) / m)
Now, let's plug in the given values and calculate the speed:
v = √((8.20 x 10^-8 N * 5.29 x 10^-11 m) / 9.11 x 10^-31 kg)
v ≈ √((4.3318 x 10^-18 N m) / 9.11 x 10^-31 kg)
v ≈ √(4.7598 x 10^12 m^2/s^2)
v ≈ 2.18 x 10^6 m/s
Therefore, the speed of the electron is approximately 2.18 x 10^6 m/s.
To find the period of the electron's orbit, we can utilize the formula:
T = (2π * r) / v
Where:
T = period of the electron (what we want to find)
r = radius of the electron's orbit (5.29 x 10^-11 m)
v = speed of the electron (2.18 x 10^6 m/s)
Now, let's substitute the values and calculate the period:
T = (2π * 5.29 x 10^-11 m) / (2.18 x 10^6 m/s)
T ≈ 1.21 x 10^-14 s
Therefore, the period of the electron's orbit is approximately 1.21 x 10^-14 seconds.