Question

伦琴射线管中阳极与阴极间所加电压为3×10⁴ V，求伦琴射线管中所发出的X射线的最短波长是多少?(电子电荷量e＝1.6×10^{－19 C}，电子的初速度可认为等于零)

Answer 1

4.1×10^{－11 m}

解析:

从阴极射出的电子经加速后射到阳极表面时具有的能量为

mv²=eU                                                                                                        (3分)

设伦琴射线的最短波长为λ，则

hν=hmv²                                                                                                                                                                                           (3分)

由以上两式解得

λ= m=4.1×10^{－11 m.                                       (3}分)