Atomic StructureHardBloom L2

Question

X-rays are produced in an X-ray tube operating at a given accelerating voltage. The wavelengths of the continuous X-ray spectrum range from:

Options

A.$0$ to $\infty$
B.$\lambda_{\min}$ to $\infty$, where $\lambda_{\min} > 0$
C.$0$ to $\lambda_{\max}$, where $\lambda_{\max}$ is finite
D.$\lambda_{\min}$ to $\lambda_{\max}$, where $0 < \lambda_{\min} < \lambda_{\max} < \infty$

Solution

{"given":"An X-ray tube is operated at accelerating voltage $V$. Electrons gain kinetic energy $eV$ and upon striking the target produce a continuous (bremsstrahlung) X-ray spectrum.","key_observation":"The maximum energy of any emitted photon equals $eV$ (complete conversion of electron kinetic energy), giving a sharp minimum wavelength $\\lambda_{\\min} = \\dfrac{hc}{eV} > 0$. Partial energy transfers produce photons of lower energy (longer wavelength), so all wavelengths from $\\lambda_{\\min}$ to $\\infty$ are present.","option_analysis":[{"label":"(A)","text":"0 to infinity","verdict":"incorrect","explanation":"Wavelengths starting from $0$ would require photon energies of $\\infty$, which is not possible. There is a definite lower cutoff $\\lambda_{\\min} = hc/eV > 0$."},{"label":"(B)","text":"λ_min to infinity, where λ_min > 0","verdict":"correct","explanation":"The minimum wavelength is $\\lambda_{\\min} = hc/eV > 0$ (Duane–Hunt limit). All longer wavelengths up to $\\infty$ are produced by partial energy-transfer collisions, making this the correct range."},{"label":"(C)","text":"0 to λ_max where λ_max is finite","verdict":"incorrect","explanation":"Starting from $0$ is impossible (requires infinite photon energy), and there is no fixed upper cutoff — wavelengths extend to $\\infty$ for vanishingly small photon energies."},{"label":"(D)","text":"λ_min to λ_max where 0 < λ_min < λ_max","verdict":"incorrect","explanation":"While $\\lambda_{\\min}$ is well-defined, there is no upper bound $\\lambda_{\\max}$ on the continuous spectrum; in principle, arbitrarily long wavelengths (very low energy photons) can be emitted."}],"answer":"(B)","formula_steps":[]}

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