Kumar Mittal Solutions for Exercise 2: For DIFFERENT COMPETITIVE EXAMINATIONS

Electrons with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-ray is:

The intensity $I$ of X-rays from the Coolidge tube is plotted against wavelength $\lambda$, as shown in the figure. The minimum wavelength found is ${\lambda }_C$ and the wavelength of the $K_{\alpha }$ line is ${\lambda }_K$. As the accelerating voltage is increased:

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:

The $K_{\alpha }$ X-ray emission line of tungsten occurs at $\lambda =0.021 \mathrm{nm}$. The energy difference between $K$ and $L$ levels in this atom is about:

The wavelength of ${\mathrm{K}}_{\mathrm{\alpha }}$ X-ray line of an element of atomic number $\mathrm{Z}=11$ is $\mathrm{\lambda }$. The wavelength of ${\mathrm{K}}_{\mathrm{\alpha }}$ X-ray line of another element of atomic number $\mathrm{Z}\text{'}$ is $4 \mathrm{\lambda }$. Then $\mathrm{Z}\text{'}$ is if $b=1$:

If ${\lambda }_{Cu}$ is the wavelength of $K_{\alpha }$ X-ray line of copper (atomic number $29$) and ${\lambda }_{\mathrm{Mo}}$ is the wavelength of the $K_{\alpha }$ X-ray line of molybdenum (atomic number $42$), then the ratio $\frac{{\lambda }_{\mathrm{Cu}}}{{\lambda }_{\mathrm{Mo}}}$ is close to: