Question

The energies for an electron in the KK, LL, and MM shells of the tungsten atom are −69,500-69,500 eV, −12,000-12,000 eV, and −2,200-2,200 eV, respectively. Calculate the wavelengths of the KαK_{\alpha} and KβK_{\beta} x rays of tungsten.

Accepted Answer

Step 1: Understand the problem. The Ka and Kb x-rays are emitted when electrons transition to the K shell from higher energy shells. Specifically, Ka corresponds to the transition from the L shell to the K shell, and Kb corresponds to the transition from the M shell to the K shell. The energy difference between these shells determines the energy of the emitted x-rays. Step 2: Calculate the energy of the Ka x-ray. Use the formula for energy difference: $$ E_{Ka} = E_{K} - E_{L} $$, where $$ E_{K} $$ and $$ E_{L} $$ are the energies of the K and L shells, respectively. Substitute the given values: $$ E_{K} = -69,500 \, 	ext{eV} $$ and $$ E_{L} = -12,000 \, 	ext{eV} . $$Step 3: Calculate the energy of the Kb x-ray. Use the formula for energy difference: $$ E_{Kb} = E_{K} - E_{M} $$, where $$ E_{K} $$ and $$ E_{M} $$ are the energies of the K and M shells, respectively. Substitute the given values: $$ E_{K} = -69,500 \, 	ext{eV} $$ and $$ E_{M} = -2200 \, 	ext{eV} . $$Step 4: Convert the x-ray energies to wavelengths using the equation $$ \lambda = \frac{hc}{E} $$, where $$ h  is $$Planck's constant $$ (6.626 	imes 10^{-34} \; 	ext{J·s}) $$, $$ c  is $$the speed of light $$ (3.00 	imes 10^{8} \; 	ext{m/s}) $$, and $$ E  is $$the energy in joules. To convert eV to joules, use the conversion factor $$ 1 \; 	ext{eV} = 1.602 	imes 10^{-19} \; 	ext{J} . $$Step 5: Perform the calculations for $$ \lambda_{Ka} $$ and $$ \lambda_{Kb} $$ using the respective energies $$ E_{Ka} $$ and $$ E_{Kb} . $$Ensure units are consistent throughout the calculation, and express the wavelengths in meters or nanometers as appropriate.

The energies for an electron in the $K$ , $L$ , and $M$ shells of the tungsten atom are $negative 69 comma 500$ eV, $negative 12 comma 000$ eV, and $negative 2 comma 200$ eV, respectively. Calculate the wavelengths of the $K_{α}$ and $K_{β}$ x rays of tungsten.

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Concetti chiave

Energy Levels in Atoms

X-ray Emission

Wavelength Calculation

The energies for an electron in the KK, LL, and MM shells of the tungsten atom are −69,500-69,500 eV, −12,000-12,000 eV, and −2,200-2,200 eV, respectively. Calculate the wavelengths of the KαK_{\(\alpha\)} and KβK_{\(\beta\)} x rays of tungsten.

Risposta video verificata per un problema simile:

Concetti chiave

Energy Levels in Atoms

X-ray Emission

Wavelength Calculation

The energies for an electron in the $K$ , $L$ , and $M$ shells of the tungsten atom are $negative 69 comma 500$ eV, $negative 12 comma 000$ eV, and $negative 2 comma 200$ eV, respectively. Calculate the wavelengths of the $K_{α}$ and $K_{β}$ x rays of tungsten.