Determine the wavelength (in nm) emitted when Be3+ ion transit from energy level n=2 to 1. ( Hint: calculate the ionization energy of Be3+, treating hydrogen-like ion the charge on the nucleus is Ze i.e. multiply Z2 in the derived expression of hydrogen energy En)
Question
Determine the wavelength (in nm) emitted when Be3+ ion transit from energy level n=2 to 1. ( Hint: calculate the ionization energy of Be3+, treating hydrogen-like ion the charge on the nucleus is Ze i.e. multiply Z2 in the derived expression of hydrogen energy En)
Solution
To solve this problem, we will use the Rydberg formula for hydrogen-like atoms:
En = -13.6 * Z^2 / n^2 eV
where Z is the atomic number, n is the energy level, and En is the energy at that level.
Step 1: Calculate the energy at n=2 and n=1 for Be3+ ion.
For Be3+, Z=4 (as it is the 4th element in the periodic table).
E1 = -13.6 * 4^2 / 1^2 = -217.6 eV E2 = -13.6 * 4^2 / 2^2 = -54.4 eV
Step 2: Calculate the energy difference between these two levels.
ΔE = E2 - E1 = -54.4 - (-217.6) = 163.2 eV
This is the energy emitted when the electron transits from n=2 to n=1.
Step 3: Convert this energy to wavelength.
We use the formula E = hc/λ, where h is Planck's constant (4.1357 x 10^-15 eV.s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength.
Rearranging for λ gives λ = hc / E.
Substituting the values gives λ = (4.1357 x 10^-15 eV.s * 3 x 10^8 m/s) / 163.2 eV = 7.57 x 10^-9 m
Step 4: Convert this to nm.
1 m = 10^9 nm, so λ = 7.57 nm.
So, the wavelength emitted when Be3+ ion transits from energy level n=2 to n=1 is approximately 7.57 nm.
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