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The average lifetime of the first excited level of a hydrogen atom is 1.0 ×10−8 s. In the Bohr model, how many orbits does an electron in the n = 2 level complete before returning to the ground level?
- a)9.7 ×106
- b)8.2 ×106
- c)9.2 ×106
- d)8.7 ×106
Correct answer is option 'B'. Can you explain this answer?
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Explanation:
In the Bohr model, the electrons in the hydrogen atom revolve around the nucleus in specific energy levels or orbits. These energy levels are represented by the principal quantum number (n). The ground state of hydrogen is represented by n=1, and the first excited state is represented by n=2.
Step 1: Find the time period (T) of one complete revolution of the electron in the n=2 level.
The average lifetime of the first excited level is given as 1.0 x 10^8 s. This is the time taken by the electron to transition from the n=2 level to the ground state. Since the electron completes one full revolution in this time, we can equate it to the time period (T) of the revolution.
T = 1.0 x 10^8 s
Step 2: Find the time taken for one complete revolution in the n=1 level.
According to the Bohr model, the energy of an electron in a particular orbit is given by E = (-13.6 eV) / n^2, where eV represents electron volts.
For the ground state (n=1), the energy is E1 = -13.6 eV.
Using the time-energy uncertainty principle, we can relate the energy of the electron to the time taken for one complete revolution:
ΔE Δt >= h/4π
ΔE = E1 - E2 (energy difference between n=1 and n=2 levels)
Δt = T (time period of revolution in the n=1 level)
h = Planck's constant
Step 3: Calculate the number of orbits completed.
Using the equation from step 2, we can solve for the number of orbits (N) completed by the electron in the n=2 level:
(N x h) / (4π) = (-13.6 eV) / ΔE
N = (-13.6 eV x 4π) / (ΔE x h)
Substituting the values, we get:
N = (-13.6 eV x 4π) / ((-13.6 eV) - (-3.4 eV)) = 8.2 x 10^6
Therefore, an electron in the n=2 level completes approximately 8.2 x 10^6 orbits before returning to the ground level.
In the Bohr model, the electrons in the hydrogen atom revolve around the nucleus in specific energy levels or orbits. These energy levels are represented by the principal quantum number (n). The ground state of hydrogen is represented by n=1, and the first excited state is represented by n=2.
Step 1: Find the time period (T) of one complete revolution of the electron in the n=2 level.
The average lifetime of the first excited level is given as 1.0 x 10^8 s. This is the time taken by the electron to transition from the n=2 level to the ground state. Since the electron completes one full revolution in this time, we can equate it to the time period (T) of the revolution.
T = 1.0 x 10^8 s
Step 2: Find the time taken for one complete revolution in the n=1 level.
According to the Bohr model, the energy of an electron in a particular orbit is given by E = (-13.6 eV) / n^2, where eV represents electron volts.
For the ground state (n=1), the energy is E1 = -13.6 eV.
Using the time-energy uncertainty principle, we can relate the energy of the electron to the time taken for one complete revolution:
ΔE Δt >= h/4π
ΔE = E1 - E2 (energy difference between n=1 and n=2 levels)
Δt = T (time period of revolution in the n=1 level)
h = Planck's constant
Step 3: Calculate the number of orbits completed.
Using the equation from step 2, we can solve for the number of orbits (N) completed by the electron in the n=2 level:
(N x h) / (4π) = (-13.6 eV) / ΔE
N = (-13.6 eV x 4π) / (ΔE x h)
Substituting the values, we get:
N = (-13.6 eV x 4π) / ((-13.6 eV) - (-3.4 eV)) = 8.2 x 10^6
Therefore, an electron in the n=2 level completes approximately 8.2 x 10^6 orbits before returning to the ground level.
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The average lifetime of the first excited level of a hydrogen atom is 1.0 ×10−8 s. In the Bohr model, how many orbits does an electron in the n = 2 level complete before returning to the ground level?a)9.7 ×106b)8.2 ×106c)9.2 ×106d)8.7 ×106Correct answer is option 'B'. Can you explain this answer?
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