The wavelength of the most energetic $X$-rays emitted when a metal target is bombarded by $40 \ keV$ electrons is approximately ....... $\mathring{A}$. $(h = 6.62 \times 10^{-34} \ J \cdot s; 1 \ eV = 1.6 \times 10^{-19} \ J; c = 3 \times 10^8 \ m/s)$

In an $X$-ray tube,electrons bombarding the target produce $X$-rays of minimum wavelength $1\; \mathring{A}$. What must be the energy of the bombarding electrons in $eV$?

If $f_1, f_2$,and $f_3$ are the frequencies of $K_\alpha, K_\beta$,and $L_\alpha$ $X$-rays of a given metal,respectively,then:

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 $hc = 12400 \ eV \mathring{A}$)

The diagram shows a graph between the intensity of $X$-rays emitted by a molybdenum target and the wavelength,when electrons of $30 \ keV$ are incident on the target. In the graph,one peak corresponds to the $K_\alpha$ line and the other peak corresponds to the $K_\beta$ line.

An $X$-ray tube operates at $50 \ kV$. It converts $1\%$ of the energy into $X$-rays. If the amount of heat produced is $495 \ W$, then the number of electrons striking the target per second is: