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InterviewSolution
| 1. |
Electrons of mass m with de-Broglie wavelength λfall on the target in an X-ray tube. The cutoffwavelength (A) of the emitted X-ray is\begin{array}{l}{\text { (1) } \lambda_{0}=\frac{2 m c \lambda^{2}}{h}} \\ {\text { (2) } \lambda_{0}=\frac{2 h}{m c}} \\ {\text { (3) } \lambda_{0}=\frac{2 m^{2} c^{2} \lambda^{3}}{h^{2}}} \\ {\text { (4) } \lambda_{0}=\lambda}\end{array} |
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