The wavelength of the ${K_\alpha}$ line for an element of atomic number $29$ is $\lambda$. Then the wavelength of the ${K_\alpha}$ line for an element of atomic number $15$ is (Take Moseley's constant $b = 1$ for both elements).

The $K_{\alpha}$ $X$-ray of molybdenum has a wavelength of $0.071 \, nm$. If the energy of a molybdenum atom with a $K$ electron knocked out is $27.5 \, keV$,the energy of this atom when an $L$ electron is knocked out will be $.... \, keV$. (Round off to the nearest integer) $[h = 4.14 \times 10^{-15} \, eVs, c = 3 \times 10^{8} \, ms^{-1}]$

The energy of the $L$ shell is $7200$ units and the energy of the $K$ shell is $1800$ units. Therefore,the energy of the $K_\alpha$ line is equal to ...... units.

The wavelength of the $K_{\alpha}$ line for an element with atomic number $57$ is $\lambda$. What will be the wavelength of the $K_{\alpha}$ line for an element with atomic number $29$?

If $10000\, V$ is applied across an $X$-ray tube,what will be the ratio of the de-Broglie wavelength of the incident electrons to the shortest wavelength of $X$-rays produced? (Given: $\frac{e}{m}$ for an electron is $1.8 \times 10^{11}\, C\, kg^{-1}$)

Two elements $A$ and $B$ with atomic numbers $Z_{A}$ and $Z_{B}$ are used to produce characteristic $X$-rays with frequencies $v_{A}$ and $v_{B}$ respectively. If $Z_{A}: Z_{B} = 1: 2$,then $v_{A}: v_{B}$ will be