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(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at $0.45$ $Å$. What is the maximum energy of a photon in the radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?
Solution:
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Verified Answer
Given: $\lambda=0.45 \mathrm{R}=0.45 \times 10^{-10} \mathrm{~m}$, $\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3 \times 10^{+8} \mathrm{~m} / \mathrm{s}$.
(a) Max. energy of photon
$$
\begin{aligned}
&\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.6 \times 10^{34} \times 3 \times 10^8}{0.45 \times 10^{-10}} \\
&=4.13 \times 10^{-15} \mathrm{~J}=\frac{4.13 \times 10^{-15} \mathrm{~J}}{1.6 \times 10^{-19}} \\
&\cong 25800 \mathrm{eV}=25.8 \mathrm{KeV}
\end{aligned}
$$
(b) To produce $\mathrm{e}^{-}$of energy $25.8 \mathrm{KeV}$, potential required is
$$
\mathrm{V}=\frac{\mathrm{eV}}{\mathrm{e}}=25.8 \mathrm{kV} \simeq 26 \mathrm{kV}
$$
(a) Max. energy of photon
$$
\begin{aligned}
&\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.6 \times 10^{34} \times 3 \times 10^8}{0.45 \times 10^{-10}} \\
&=4.13 \times 10^{-15} \mathrm{~J}=\frac{4.13 \times 10^{-15} \mathrm{~J}}{1.6 \times 10^{-19}} \\
&\cong 25800 \mathrm{eV}=25.8 \mathrm{KeV}
\end{aligned}
$$
(b) To produce $\mathrm{e}^{-}$of energy $25.8 \mathrm{KeV}$, potential required is
$$
\mathrm{V}=\frac{\mathrm{eV}}{\mathrm{e}}=25.8 \mathrm{kV} \simeq 26 \mathrm{kV}
$$
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