The wavelength of the most energetic X-rays e | vedclass

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(D) The most energetic $X$-ray photon corresponds to the case where the entire kinetic energy of the electron is converted into a single photon.$E = K

Vedclass · Accepted Answer

$0.31$

Vedclass · Answer

(D) The most energetic $X$-ray photon corresponds to the case where the entire kinetic energy of the electron is converted into a single photon.$E = K.E. = 40 \ keV = 40 	imes 10^3 \ eV$.The energy of a photon is given by $E = \frac{hc}{\lambda}$.Thus,$\lambda = \frac{hc}{E}$.Using the relation $\lambda (	ext{in } \mathring{A}) \approx \frac{12400}{E (	ext{in } eV)}$,we get:$\lambda = \frac{12400}{40000} \mathring{A} = 0.31 \mathring{A}$.Alternatively,using constants: $\lambda = \frac{(6.62 	imes 10^{-34} \ J \cdot s) 	imes (3 	imes 10^8 \ m/s)}{40 	imes 10^3 	imes 1.6 	imes 10^{-19} \ J} = 0.309 	imes 10^{-10} \ m = 0.309 \ \mathring{A} \approx 0.31 \ \mathring{A}$.

The wavelength of the most energetic $X$-rays emitted when a metal target is bombarded by $40 \ keV$ electrons is approximately ....... $\mathring{A}$. $(h = 6.62 \times 10^{-34} \ J \cdot s; 1 \ eV = 1.6 \times 10^{-19} \ J; c = 3 \times 10^8 \ m/s)$

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