Analyzing the Setup
Imagine a hydrogen-like atom. We have a positively charged nucleus at the center, and an electron revolving around it in some excited state. The problem gives us two crucial pieces of information about this electron: its orbital angular momentum and the radius of its orbit.
Our first goal is to identify the exact state of the electron and the identity of the atom itself.
The Master Equations
According to Bohr's quantization condition, the orbital angular momentum
L of an electron in the
n-th orbit is an integral multiple of
2πh:
L=2πnh
We are given that
L=2π3h. Comparing this with Bohr's formula, we can immediately deduce the principal quantum number:
n=3
So, our electron is sitting in the third orbit!
Next, we use the formula for the radius of the
n-th orbit in a hydrogen-like atom:
rn=Zn2a0
where
a0 is the Bohr radius and
Z is the atomic number.
We are given that the radius is
4.5a0. Let's substitute
n=3 into our radius formula:
4.5a0=Z32a0
4.5=Z9
Z=4.59=2
This is a fantastic revelation! An atomic number of Z=2 means our atom is actually a singly ionized Helium atom (He+).
Calculating the Wavelengths
Now, the atom de-excites. The electron can jump from its current state (
n=3) to lower energy levels (
n=2 or
n=1), emitting photons in the process. The wavelength
λ of these emitted photons is given by the Rydberg formula:
λ1=RZ2(nf21−ni21)
Let's calculate the wavelengths for all possible transitions.
Transition 1: From n=3 to n=2
λ11=R(2)2(221−321)
λ11=4R(41−91)=4R(365)=95R
λ1=5R9
Transition 2: From n=3 to n=1
λ21=R(2)2(121−321)
λ21=4R(1−91)=4R(98)=932R
λ2=32R9
Transition 3: From n=2 to n=1
If the electron first jumped to
n=2, it will subsequently jump to
n=1.
λ31=R(2)2(121−221)
λ31=4R(1−41)=4R(43)=3R
λ3=3R1
Final Conclusion
The possible wavelengths emitted during the de-excitation process are 5R9, 32R9, and 3R1.
Comparing these results with the given options, we find that 32R9 corresponds to option (a) and 5R9 corresponds to option (c). Therefore, the correct options are (a) and (c).
