Question

For a given material, the energy $\left(E\right)$ and wavelength $\left(\lambda \right)$ of characteristic $\text{X}$-rays satisfies

• A. $E\left(K_{\alpha }\right)>E\left(K_{\beta }\right)>E\left(K_{\gamma }\right)$
• B. $E\left(M_{\alpha }\right)>E\left(L_{\alpha }\right)>E\left(K_{\alpha }\right)$
• C. $\lambda \left(K_{\alpha }\right)>\lambda \left(K_{\beta }\right)>\lambda \left(K_{\gamma }\right)$
• D. $\lambda \left(M_{\alpha }\right)>\lambda \left(L_{\alpha }\right)>\lambda \left(K_{\gamma }\right)$

Answer 1

Answer: (C), (D)

$\lambda \left(M_{\alpha }\right)>\lambda \left(L_{\alpha }\right)>\lambda \left(K_{\gamma }\right)$

For characteristic $\text{X}$- ray,


From the above diagram,

$E\left(K_{\alpha }\right)<E\left(K_{\beta }\right)<E\left(K_{\gamma }\right)$

As we know that, $E\propto \frac{1}{\lambda }$

Hence, $\lambda \left(K_a\right)>\lambda \left(K_b\right)>\lambda \left(K_{\gamma }\right)$

Also, $E\left(M_{\alpha }\right)<E\left(L_{\alpha }\right)<E\left(K_{\alpha }\right)$

$\Rightarrow \lambda \left(M_{\alpha }\right)>\lambda \left(L_{\alpha }\right)>\lambda \left(K_{\alpha }\right)$

Therefore, options $\left(\text{C}\right)$ and $\left(\text{D}\right)$ are the correct answer.