The potential difference between the cathode and the target in a Coolidge tube is $100 \ kV$. The minimum wavelength of the $X$-rays emitted by the tube is

$\Delta \lambda$ is the difference between the wavelength of the $K_\alpha$ line and the minimum wavelength of the continuous $X$-ray spectrum when the $X$-ray tube is operated at a voltage $V$. If the operating voltage is changed to $V / 3$,then the above difference is $\Delta \lambda^{\prime}$. Then:

$X$-rays of wavelength $0.140 \,nm$ are scattered from a block of carbon. What will be the wavelengths of $X$-rays scattered at $90^{\circ}$ (in $\,nm$)?

$(a)$ An $X$-ray tube produces a continuous spectrum of radiation with its short wavelength end at $0.45\,\mathring{A}$. 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?

The wavelength of ${K_{\alpha }}$ $X$-rays produced by an $X$-ray tube is $0.76 \, \mathring{A}$. The atomic number of the anticathode material is:

The dependence of the short wavelength limit $\lambda _{\min }$ on the accelerating potential $V$ is represented by the curve of figure.