An electron drops from the L shell to the K shell and gives off an X-ray with a wavelength of 0.0205 nm.
What is the atomic number of this atom?
If the atom were hydrogen, you would be getting the Lyman alpha line, with a wavelength of 121.6 nm. The observed wavelength is 5932 times less. Take the square toot of that for the atomic number, which would be 77.
I meant square root, not toot.
To determine the atomic number of the atom, we can use the relationship between the energy levels of electrons in an atom and the emitted wavelength of X-rays.
The energy difference between the L and K shells can be calculated using the equation:
ΔE = hc / λ
Where:
ΔE is the energy difference between the L and K shells,
h is Planck's constant (6.626 × 10^-34 J·s),
c is the speed of light (3.00 × 10^8 m/s), and
λ is the wavelength of the X-ray (0.0205 nm, which can be converted to meters by multiplying by 10^-9).
Substituting the given values into the equation:
ΔE = (6.626 × 10^-34 J·s)(3.00 × 10^8 m/s) / (0.0205 × 10^-9 m)
= 9.66 × 10^-16 J
Next, we can use the equation for the energy difference between electron energy levels in an atom:
ΔE = -13.6 eV * (Z^2 / n^2)
Where:
ΔE is the energy difference between the shells (9.66 × 10^-16 J),
-13.6 eV is the energy of an electron at the n-th energy level in an atom,
Z is the atomic number (the value we're trying to find), and
n is the energy level of the electron (the initial energy level, K shell, in this case).
Rearranging the equation to solve for Z:
Z = sqrt(-ΔE * n^2 / -13.6 eV)
Substituting the values:
Z = sqrt(-(9.66 × 10^-16 J) * (1^2) / -13.6 eV)
= sqrt(7.12 × 10^-16)
≈ 2.67
Since the atomic number must be a whole number, rounding 2.67 to the nearest whole number gives us the atomic number of this atom, which is 3.
Therefore, the atomic number of this atom is 3.