Atoms And Nuclei Question 280
Question: Suppose that a material emits X-rays of wavelengths $ \lambda {\kappa_{\alpha }},\lambda {\kappa_{\beta }},{\lambda_{{L_{\alpha }}}} $ , when it is excited by fast moving electrons; the wavelengths corresponding to $ {K_{\alpha }},{K_{\beta }},{L_{\alpha }} $ X- rays of the material respectively. Then we can write
Options:
A) $ \lambda {\kappa_{\beta }}=\lambda {\kappa_{\alpha }}+{\lambda_{{L_{\alpha }}}} $
B) $ \sqrt{\lambda {\kappa_{\beta }}}=\sqrt{\lambda {\kappa_{\alpha }}}+\sqrt{{\lambda_{{L_{\alpha }}}}} $
C) $ \frac{1}{\lambda {\kappa_{\beta }}}=\frac{1}{\lambda {\kappa_{\alpha }}}+\frac{1}{{\lambda_{{L_{\alpha }}}}} $
D) $ \frac{1}{\sqrt{\lambda {\kappa_{\beta }}}}=\frac{1}{\sqrt{\lambda {\kappa_{\alpha }}}}+\frac{1}{\sqrt{{\lambda_{{L_{\alpha }}}}}} $
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Answer:
Correct Answer: C
Solution:
- The energy level diagram of the atom is . It is clear that $ {E_{{\kappa_{\beta }}}}={E_{\kappa \alpha }}+{E_{{L_{\alpha }}}} $ $ or,{v_{{\kappa_{\beta }}}}={v_{\kappa \alpha }}+{v_{{L_{\alpha }}}} $ $ or,\frac{1}{{\lambda_{{\kappa_{\beta }}}}}=\frac{1}{{\lambda_{\kappa \alpha }}}+\frac{1}{{\lambda_{{L_{\alpha }}}}} $
BETA
