Electrons of mass $m$ with de-Broglie wavelength $\lambda$ fall on the target in an $X$-ray tube. The cutoff wavelength $(\lambda_0)$ of the emitted $X$-ray is

When $0.50\; \mathring{A}$ $X$-rays strike a material,the photoelectrons from the $K$-shell are observed to move in a circle of radius $23\; mm$ in a magnetic field of $2 \times 10^{-2}\; T$ acting perpendicularly to the direction of emission of photoelectrons. The binding energy of $K$-shell electrons is $..........\; keV$.

The difference between $\lambda_{K_{\alpha}}$ and $\lambda_{min}$ becomes twice if the accelerating potential changes from $3100 \ V$ to $12400 \ V$. Find $\lambda_{K_{\alpha}}$ in $\mathring{A}$. (Given $hc = 12400 \ eV \mathring{A}$)

An $X$-ray tube with a copper target emits a $Cu$ $K_{\alpha}$ line of wavelength $1.50 \ \mathring{A}$. What should be the minimum voltage through which electrons are to be accelerated to produce this wavelength of $X$-rays? $(h = 6.63 \times 10^{-34} \ \text{J}\cdot\text{s}, c = 3 \times 10^8 \ \text{m/s})$

In an $X-ray$ tube,the intensity of the emitted $X-ray$ beam is increased by

The typical wavelength of $X$-ray is