The radius of circular electron orbits in the Bohr model of the hydrogen atom are given by (5.29 ✕ 10^−11 m)n^2, where n is the electron's energy level (see figure below). The speed of the electron in each energy level is (c/137n), where c = 3 ✕ 10^8 m/s is the speed of light in vacuum.(a) What is the centripetal acceleration of an electron in the ground state (n = 1)
of the Bohr hydrogen atom? m/^2 b) What are the magnitude and direction of the centripetal force acting on an electron in the ground state?N(c) What are the magnitude and direction of the centripetal force acting on an electron in the
n = 2
excited state? N
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To calculate the centripetal acceleration, centripetal force, and their magnitudes and directions, we will use the following formulas:
1. Centripetal acceleration (aᵣ):
aᵣ = v²/r
where v is the velocity of the electron and r is the radius of its circular orbit.
2. Centripetal force (Fᵣ):
Fᵣ = m * aᵣ
where m is the mass of the electron.
Let's solve the given parts step by step:
(a) What is the centripetal acceleration of an electron in the ground state (n = 1) of the Bohr hydrogen atom?
To find the centripetal acceleration (aᵣ) in the ground state, we need to calculate the velocity (v) and radius (r) using the given formulas.
Given:
Radius (r) = (5.29 × 10^(-11) m) * n²
Velocity (v) = c / (137n)
c = 3 × 10^8 m/s (speed of light in vacuum)
Substituting the values into the formulas, we have:
r = (5.29 × 10^(-11) m) * 1² = 5.29 × 10^(-11) m
v = (3 × 10^8 m/s) / (137 × 1) = 2.19 × 10^6 m/s
Now, we can calculate the centripetal acceleration using the formula:
aᵣ = v²/r = (2.19 × 10^6 m/s)² / (5.29 × 10^(-11) m) ≈ 9.02 × 10^21 m/s²
Therefore, the centripetal acceleration of an electron in the ground state of the Bohr hydrogen atom is approximately 9.02 × 10^21 m/s².
(b) What are the magnitude and direction of the centripetal force acting on an electron in the ground state?
To find the magnitude and direction of the centripetal force (Fᵣ) in the ground state, we need to calculate the mass of the electron (m) from known values.
Given:
Mass of the electron (m) = 9.11 × 10^(-31) kg
Now we can calculate the centripetal force using the formula:
Fᵣ = m * aᵣ = (9.11 × 10^(-31) kg) * (9.02 × 10^21 m/s²) ≈ 8.24 × 10^(-9) N
The magnitude of the centripetal force acting on the electron in the ground state is approximately 8.24 × 10^(-9) N. The direction of the force is towards the center of the circular orbit.
(c) What are the magnitude and direction of the centripetal force acting on an electron in the n = 2 excited state?
To find the centripetal force (Fᵣ) in the n = 2 excited state, we can use the same mass of the electron (m) and the given formulas to calculate the radius (r) and velocity (v).
Given:
Radius (r) = (5.29 × 10^(-11) m) * (2)² = 2.12 × 10^(-10) m
Velocity (v) = (3 × 10^8 m/s) / (137 × 2) = 1.09 × 10^6 m/s
Now, we can calculate the centripetal acceleration using the formula:
aᵣ = v²/r = (1.09 × 10^6 m/s)² / (2.12 × 10^(-10) m) ≈ 5.57 × 10^20 m/s²
Next, we can calculate the centripetal force using the formula:
Fᵣ = m * aᵣ = (9.11 × 10^(-31) kg) * (5.57 × 10^20 m/s²) ≈ 5.08 × 10^(-10) N
Therefore, the magnitude of the centripetal force acting on an electron in the n = 2 excited state is approximately 5.08 × 10^(-10) N. The direction of the force is towards the center of the circular orbit.