In tungsten,the energy of a K-shell electron | vedclass

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(B) The energy of the emitted photon $(E)$ is equal to the difference in energy between the two shells: $E = E_K - E_L$. Given $E_K = 20 \, keV$ and $

Vedclass · Accepted Answer

$0.6887 \, \mathring{A}$

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(B) The energy of the emitted photon $(E)$ is equal to the difference in energy between the two shells: $E = E_K - E_L$. Given $E_K = 20 \, keV$ and $E_L = 2 \, keV$,the energy difference is $\Delta E = 20 \, keV - 2 \, keV = 18 \, keV = 18000 \, eV$. The wavelength $\lambda$ is given by the formula $\lambda = \frac{hc}{E}$. Using the relation $\lambda \approx \frac{12400 \, eV \cdot \mathring{A}}{E 	ext{ (in } eV)}$,we get: $\lambda = \frac{12400}{18000} \, \mathring{A} = 0.6888 \, \mathring{A}$. Rounding to the nearest option,the wavelength is $0.6887 \, \mathring{A}$.

In tungsten,the energy of a $K$-shell electron is $20 \, keV$ and that of an $L$-shell electron is $2 \, keV$. When an electron jumps from the $L$ shell to the $K$ shell,the wavelength of the emitted $X$-ray photon will be:

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