The ${K_{\alpha}}$ $X$-rays arising from a cobalt $(Z = 27)$ target have a wavelength of $179\, pm$. The ${K_{\alpha}}$ $X$-rays arising from a nickel target $(Z = 28)$ is

$X$-rays of wavelength $0.1 \ \mathring A$ are allowed to fall on a metal and get scattered. The wavelength of the scattered radiation is $0.111 \ \mathring A$. If $h = 6.624 \times 10^{-34} \ J \cdot s$ and $m_0 = 9 \times 10^{-31} \ kg$,then the direction of the scattered photons will be:

Let ${\lambda _\alpha }$,${\lambda _\beta }$ and ${\lambda '_\alpha }$ denote the wavelengths of the $X-$ rays of the ${K_\alpha }, {K_\beta }$ and ${L_\alpha }$ lines in the characteristic $X-$ rays for a metal.

An $X$-ray machine has an accelerating potential difference of $25,000$ volts. By calculation, the shortest wavelength will be obtained as....... $\mathring{A}$ ($h = 6.62 \times 10^{-34} \text{ J-s}$; $e = 1.6 \times 10^{-19} \text{ C}$).

$\Delta \lambda$ 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:

The ratio of the energy of an $X$-ray photon of wavelength $1 \mathring{A}$ to that of visible light of wavelength $5000 \mathring{A}$ is: