If the attenuation coefficient of aluminum for soft $X$-rays is $1.73 \, cm^{-1}$,what percentage of $X$-rays will pass through an aluminum sheet of thickness $1.156 \, cm$?

An electron having de-Broglie wavelength $\lambda$ is incident on a target in an $X$-ray tube. The cut-off wavelength of the emitted $X$-ray is:

$A$ potential difference of $42,000 \ V$ is used in an $X-ray$ tube to accelerate electrons. The maximum frequency of the $X-radiations$ produced is ($1 \ eV = 1.6 \times 10^{-19} \ J$ and $h = 6.63 \times 10^{-34} \ J \cdot s$).

Electrons of mass $m$ with de-Broglie wavelength $\lambda$ fall on a target. The cut-off wavelength $\lambda_0$ is equal to [$h=$ Planck's constant,$c=$ velocity of light].

What $kV$ potential is to be applied on an $X$-ray tube so that the minimum wavelength of emitted $X$-rays may be $1 \text{ Å}$? $(h = 6.625 \times 10^{-34} \text{ J-s})$

The wavelength of an $X$-ray photon is $0.01 \ \mathring A$. Its momentum in $kg \ m/s$ is .........