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(B) $1$. In the characteristic $X$-ray spectrum,the $K_\alpha$ line has a longer wavelength and higher intensity compared to the $K_\beta$ line.$2$. F

Vedclass · Accepted Answer

The highest peak is the $K_\alpha$ line at $0.7 \ \mathring{A}$.

Vedclass · Answer

(B) $1$. In the characteristic $X$-ray spectrum,the $K_\alpha$ line has a longer wavelength and higher intensity compared to the $K_\beta$ line.$2$. From the graph,the peak at $0.7 \ \mathring{A}$ is higher than the peak at $0.6 \ \mathring{A}$. Therefore,the $K_\alpha$ line corresponds to the peak at $0.7 \ \mathring{A}$.$3$. If the energy of the incident electrons is increased,the minimum wavelength (cut-off wavelength) $\lambda_{min} = \frac{hc}{E}$ decreases,shifting the continuous spectrum towards the left. However,the characteristic peaks $(K_\alpha, K_\beta)$ depend only on the target material and remain at their fixed wavelengths. Thus,option $(c)$ is incorrect as the characteristic peaks do not shift,and option $(b)$ is the correct observation from the graph.

The diagram shows a graph between the intensity of $X$-rays emitted by a molybdenum target and the wavelength,when electrons of $30 \ keV$ are incident on the target. In the graph,one peak corresponds to the $K_\alpha$ line and the other peak corresponds to the $K_\beta$ line.

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