Question

Consider a metal used to produce some characteristic x-rays. Energy of x-rays is given by E and wavelength is represented by $\lambda$. Then which of the following is true.

• 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)

The correct options are
C $\lambda \left(K_{\alpha }\right)>\lambda \left(K_{\beta }\right)>\lambda \left(K_{\gamma }\right)$
D $\lambda \left(M_{\alpha }\right)>\lambda \left(L_{\alpha }\right)>\lambda \left(K_{\alpha }\right)$

From the energy diagram, $E\left(K_{\gamma }\right)>E\left(K_{\beta }\right)>E\left(K_{\alpha }\right)$
As $E\propto \frac{1}{\lambda }$, $\lambda \left(K_{\alpha }\right)>\lambda \left(K_{\beta }\right)>\lambda \left(K_{\gamma }\right)$
Also, $E\left(K_{\alpha }\right)>E\left(L_{\alpha }\right)>E\left(M_{\alpha }\right)$ and so $\lambda \left(M_{\alpha }\right)>\lambda \left(L_{\alpha }\right)>\lambda \left(K_{\alpha }\right)$