An atom is ionized so that it has only one electron. When the electron drops from the n = 6 shell to the n = 1 shell, an X-ray of wavelength 193.76 pm is emitted. What is the atomic number?
To determine the atomic number, we need to first understand the concept of electron shells and energy levels. The electron shells are labeled with principal quantum numbers (n), where n = 1 corresponds to the innermost shell and n = 2, 3, 4, and so on, represent progressively higher energy levels.
In this scenario, we have an atom that is ionized, meaning it has lost all but one electron. The electron transition mentioned is from the n = 6 shell to the n = 1 shell, resulting in the emission of an X-ray with a wavelength of 193.76 pm.
The energy of an electron in a particular shell is given by the formula:
E = -(13.6 eV)/n^2
Here, eV represents electron volts, and n is the principal quantum number. Since we're considering the transition from the n = 6 shell to the n = 1 shell, we can calculate the difference in energy levels.
First, for the n = 1 shell:
E1 = -(13.6 eV)/(1^2) = -13.6 eV
Next, for the n = 6 shell:
E2 = -(13.6 eV)/(6^2) = -0.378 eV
The energy difference ΔE between the two shells is given by:
ΔE = E2 - E1
ΔE = (-0.378 eV) - (-13.6 eV) = 13.222 eV
Now, we can convert this energy difference to wavelength using the equation:
ΔE = hc/λ
Where h is Planck's constant (6.626 x 10^-34 J.s) and c is the speed of light (3.0 x 10^8 m/s)
Converting eV to joules:
1 eV = 1.602 x 10^-19 J
So, 13.222 eV = 13.222 x 1.602 x 10^-19 J = 2.117 x 10^-18 J
Rearranging the equation to solve for wavelength (λ):
λ = hc/ΔE
Substituting the values:
λ = (6.626 x 10^-34 J.s)(3.0 x 10^8 m/s)/(2.117 x 10^-18 J) = 9.882 x 10^-12 m
Now, we can convert this wavelength from meters to picometers:
1 meter = 10^12 picometers
So, 9.882 x 10^-12 m = 9.882 x 10^0 pm = 9.882 pm
Comparing this calculated wavelength of 9.882 pm with the given wavelength of 193.76 pm, we observe a significant discrepancy. It suggests that there might be an error in the problem statement or calculations.
Hence, we cannot deduce the atomic number with the given information.