The hydrogen atomic emission spectrum includes a UV line with a wavelength of 92.3 nm.
Is this line associated with a transition between different excited states or between an excited state and the ground state?
What is the value of n1 of this transition?
What is the wavelength of the longest- wavelength photon that a ground-state hydrogen can absorb?
Oh boy, we're going into atomic stuff now! Hold onto your electrons!
To answer your questions, the UV line with a wavelength of 92.3 nm in the hydrogen atomic emission spectrum is associated with a transition between different excited states.
Now, let's move on to the value of n1 for this transition. The energy levels in hydrogen are given by the equation E = -13.6/n^2 eV, where n is the principal quantum number. To find n1, we can use the relation 1/λ = R(1/n1^2 - 1/n2^2), where λ is the wavelength, R is the Rydberg constant, and n2 is the principal quantum number of the excited state.
For the UV line with a wavelength of 92.3 nm, let's convert it to meters first. That gives us a wavelength of 92.3 × 10^(-9) m. Plugging this value into the equation, we can solve for n1.
Lastly, let's find the wavelength of the longest-wavelength photon that a ground-state hydrogen can absorb. Remember, absorption means an electron moving up from the ground state. In this case, n1 would be equal to 1, and we need to find the wavelength.
Phew, that was a lot of electron gymnastics! I hope I didn't confuse you more. Let me know if you need any more help or if you want to hear more atomic jokes!
To determine whether the UV line in the hydrogen atomic emission spectrum is associated with a transition between different excited states or between an excited state and the ground state, we need to consider the energy levels involved.
In the hydrogen atom, when an electron transitions from a higher energy level to a lower energy level, it releases energy in the form of electromagnetic radiation. This emitted radiation corresponds to specific wavelengths in the atomic emission spectrum.
The energy levels in the hydrogen atom are given by the formula:
E = -13.6 eV / n^2
where E is the energy level, -13.6 eV is the ionization energy of hydrogen, and n is an integer representing the principal quantum number.
UV radiation corresponds to short wavelengths, indicating higher energy levels. The fact that the UV line has a wavelength of 92.3 nm suggests that it is associated with a transition between two different excited states rather than between an excited state and the ground state.
To determine the value of n1, the principal quantum number of the energy level from which the transition originates, we can use the Rydberg formula:
1/λ = R_H * (1/n1^2 - 1/n2^2)
where λ is the wavelength of the emitted photon (in meters), R_H is the Rydberg constant for hydrogen (approximately 1.097 × 10^7 m^-1), and n1 and n2 are the principal quantum numbers corresponding to the initial and final energy levels, respectively.
Plugging in the values, we get:
1/92.3 nm = 1.097 × 10^7 m^-1 * (1/n1^2 - 1/2^2)
Simplifying and solving for n1, we find:
n1^2 = (1/(1.097 × 10^7 m^-1 * 1/92.3 nm) + 1/2^2)^-1
n1^2 = (9.370 × 10^6 + 0.25)^-1
n1 ≈ √(1/(9.370 × 10^6 + 0.25))
Calculating this, we find n1 ≈ 3. Therefore, the value of n1 for this transition is approximately 3.
To determine the longest-wavelength photon that a ground-state hydrogen atom can absorb, we need to consider the energy required for the electron to transition from the ground state (n=1) to a higher energy level.
Using the formula mentioned earlier:
E = -13.6 eV / n^2
we can calculate the energy difference between the ground state and various excited states.
To find the longest-wavelength photon, we need to find the minimum energy difference, which will correspond to the maximum wavelength.
The energy difference between the ground state (n=1) and an excited state (n=n2) is given by:
ΔE = |E_2 - E_1| = |-13.6 eV / n2^2 - (-13.6 eV / 1^2)|
Setting this energy difference equal to the energy of a photon (E = hf) with the longest wavelength, we can solve for n2:
|h * c / λ_max| = |-13.6 eV / n2^2 - (-13.6 eV / 1^2)|
Note: h is Planck's constant, c is the speed of light, and λ_max is the longest wavelength.
Simplifying and solving for n2, we get:
|(-6.63 × 10^-34 J s) * (3.00 × 10^8 m/s) / λ_max| = |-13.6 eV / n2^2 + 13.6 eV|
Solving this equation will give us the value of n2 for the longest-wavelength photon that the ground-state hydrogen atom can absorb. However, without knowing λ_max, we cannot determine a specific value for n2.