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Step-by-Step Solution
Step 1: Identify the relevant energy relation for Kα X-ray
The Kα X-ray photon is emitted when an electron from the L-shell fills the vacancy in the K-shell. The corresponding energy is the difference between the energy of the atom with a knocked-out K electron and the energy of the atom with a knocked-out L electron. Symbolically:
E_{k_\alpha} = E_{k} - E_{L}
Step 2: Express the photon energy in terms of wavelength
The energy of a photon (Kα X-ray) is given by:
E_{k_\alpha} = \frac{hc}{\lambda_{k_\alpha}}
where:
h is Planck's constant ( 4.14 \times 10^{-15}\,\text{eV·s} )
c is the speed of light ( 3 \times 10^8\,\text{m/s} )
\lambda_{k_\alpha} is the wavelength of the Kα X-ray ( 0.071 \,\text{nm} )
Step 3: Calculate the photon energy in keV
First convert the wavelength to meters: 0.071\,\text{nm} = 0.071 \times 10^{-9}\,\text{m}.
Now compute E_{k_\alpha} = \frac{hc}{\lambda_{k_\alpha}} in eV (and then in keV):
hc \approx 12.42 \times 10^{-7}\,\text{eV·m}
(This value is obtained by multiplying 4.14 × 10−15 eV·s by 3 × 108 m/s).
E_{k_\alpha} = \frac{12.42 \times 10^{-7}\,\text{eV·m}}{0.071 \times 10^{-9}\,\text{m}}.
Carrying out the division and converting to keV gives approximately 17.5\,\text{keV}.
Step 4: Find the energy of the atom when an L electron is knocked out
We know E_k = 27.5\,\text{keV} (energy of the atom when a K electron is knocked out). Using the relation:
E_{L} = E_{k} - E_{k_\alpha},
we get:
E_{L} = 27.5\,\text{keV} - 17.5\,\text{keV} = 10\,\text{keV}.
Final Answer
The energy of the molybdenum atom when an L electron is knocked out is 10\,\text{keV} .
