At the time of a total solar eclipse,the spectrum of solar radiation would be:

An electron beam is accelerated by a potential difference $V$ to hit a metallic target to produce $X$-rays. It produces continuous as well as characteristic $X$-rays. If $\lambda_{\min}$ is the smallest possible wavelength of $X$-ray in the spectrum,the variation of $\log \lambda_{\min}$ with $\log V$ is correctly represented in:

$X$-rays of wavelength $\lambda$ are reflected from parallel planes of atoms in a crystal with spacing $d$ between two planes, as shown in the figure. If the two reflected beams interfere constructively, then the condition for maxima will be (where $n$ is the order of the interference fringe):

The energy of an $X$-ray photon of wavelength $1.65 \mathring{A}$ is ....... $keV$. $(h = 6.6 \times 10^{-34} \ J \cdot s, \ c = 3 \times 10^8 \ m/s, \ 1 \ eV = 1.6 \times 10^{-19} \ J)$ (in $.5$)

If the wavelength of the $K_\alpha$ line of an element with atomic number $43$ is $\lambda$,then the wavelength of the $K_\alpha$ line of an element with atomic number $29$ is:

The energy of $X$-rays with a wavelength of $1.65 \, \mathring{A}$ is ............. $keV$ (Given: $h = 6.6 \times 10^{-34} \, J \cdot s$,$c = 3 \times 10^{8} \, m/s$,$1 \, eV = 1.6 \times 10^{-19} \, J$). (in $.5$)