Question

For a given material, the energy and wavelength of characteristic x-rays satisfy:

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

Answer 1

Answer: (C), (D)

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


Quantitatively,
$E\left(K_{\alpha }\right)=E_K-E_L$
$E\left(K_{\beta }\right)=E_K-E_M$
$E\left(K_{\gamma }\right)=E_K-E_N$
Clearly,
$E\left(K_{\alpha }\right)$ < $E\left(K_{\beta }\right)$ < $E\left(K_{\gamma }\right).$ (1)
Since energy is inversely proportional to wavelength, by the relation E = $\frac{hc}{\lambda }$, relation (1) will imply,
$\lambda \left(K_{\alpha }\right)$ > $\lambda \left(K_{\beta }\right)$ < $\lambda \left(K_{\gamma }\right).$ (2)
What now about the L and M lines? The diffference between the energy levels decreases as the principal quantum number n increases. This will mean -
$(E_M-E_N)$ < $(E_L-E_M)$ < $(E_K-E_L)$
$\to$ $E\left(M_{\alpha }\right)<E\left(L_{\alpha }\right)<E\left(K_{\alpha }\right)$ (3)
$\to$ $\lambda \left(M_{\alpha }\right)$ > $\lambda \left(L_{\alpha }\right)$ < $\lambda \left(K_{\alpha }\right)$ (4)
Since energy and wavelength share a reciprocal relation.