Thank you drwls,
My problem is
a. The lifetime of the levels in a hydrogen atom are of the order 10-E8s. What is the evergy uncertainty of these states? Compare this result with the energy value of the first excited state.
b. Find the shortest and the longgest wavelengths of the Lyman series of singly ionized helium.
c. In the Balmer exlperiment the wavelengths of the observed lines are 656m,48nm,434nm. Takiing the agular range of observation to be between 30 degree and 990 degree, how many orders could you see?
I solve and b I found the E uncertainty = 6.6.10-E8eV, it is greater than E of the first state. The ahortest and longest wavelength I solve is 22.8 and 30.4 but I am not sure if I right.
Question c is not required for lab report, it is the question outside the lectures I learned.
Hi bun,
I made this a take home physics final which is supposed to judge your skill level in physics 214 not drwls's skill level.
Thanks
To solve part (a) of the problem, which asks for the energy uncertainty of the hydrogen atom states, you can use the Heisenberg uncertainty principle. The energy uncertainty (ΔE) can be related to the lifetime (τ) of the states by the equation:
ΔE ≈ h/4πτ
Where h is the Planck's constant (6.626 x 10^-34 J s) and τ is the lifetime of the levels in seconds (10^-8 s in this case).
Calculating this, you get:
ΔE ≈ (6.626 x 10^-34 J s) / (4π x 10^-8 s)
Simplifying the expression, we get:
ΔE ≈ 5.27 x 10^-26 J
So the energy uncertainty of these states is approximately 5.27 x 10^-26 J.
To compare this result with the energy value of the first excited state, you need to know the energy value of the first excited state. If you have that value, you can compare it directly to the energy uncertainty value obtained above.
Moving on to part (b), where you are asked to find the shortest and longest wavelengths of the Lyman series of singly ionized helium. The Lyman series corresponds to transitions of the electron from higher energy levels to the first energy level in a hydrogen-like atom.
The shortest wavelength (λ_shortest) of the Lyman series can be calculated using the Rydberg formula:
1/λ_shortest = R_Helium/(1^2 - 2^2)
Where R_Helium is the Rydberg constant for helium. For singly ionized helium, R_Helium is approximately equal to twice the Rydberg constant for hydrogen, which is 109,677 cm^-1.
Calculating this using the given formula, you get:
1/λ_shortest = (2 x 109,677 cm^-1) / (1 - 4)
Simplifying the expression, you get:
1/λ_shortest = -219,354 cm^-1
Taking the reciprocal of both sides gives:
λ_shortest = -1/219,354 cm^-1
Perform a similar calculation for the longest wavelength (λ_longest), but now the transition is from infinity to the first energy level:
1/λ_longest = R_Helium/(1^2 - ∞^2)
Calculating this, you get:
1/λ_longest = (2 x 109,677 cm^-1) / (1 - 0)
Simplifying the expression:
1/λ_longest = 2 x 109,677 cm^-1
Taking the reciprocal gives:
λ_longest = 1/(2 x 109,677) cm^-1
You can verify these values by converting them to the appropriate units (such as nanometers) if needed.
Regarding part (c) of the question, you are asked to determine the number of orders that could be seen in the Balmer experiment given the observed wavelengths and angular range of observation.
To calculate the number of orders, you need to know the wavelength range of each order, which is given by the formula:
Δλ = λ / N
Where N is the order number and λ is the observed wavelength.
For each observed wavelength, you can calculate the range of wavelengths covered by each order using this formula. Then, consider the angular range of observation (30° to 990°) and determine how many orders fall within that range.
You should be able to apply this method to determine the number of orders seen in the Balmer experiment.