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

$\Delta \lambda$ 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 maximum wavelength of hard $X$-rays required for the study of bone fractures is $10^{-11} \ m$. The accelerating voltage in the $X$-ray machine must be:

When $X$-rays from a Coolidge tube pass through an aluminum foil of thickness $0.3 \, mm$,$50\%$ of the $X$-rays are transmitted. If the potential difference between the target and the cathode is increased,the fraction of $X$-rays transmitted through the same foil will be .......

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:

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: