The difference between lambda_{K_{alpha}} and | vedclass

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(C) The minimum wavelength of $X$-rays is given by $\lambda_{min} = \frac{hc}{V}$.Given $hc = 12400 \ eV \mathring{A}$,so $\lambda_{min} = \frac{12400

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$7$

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(C) The minimum wavelength of $X$-rays is given by $\lambda_{min} = \frac{hc}{V}$.Given $hc = 12400 \ eV \mathring{A}$,so $\lambda_{min} = \frac{12400}{V} \mathring{A}$.Let $\lambda_{K_{\alpha}} = \lambda_0$ (constant for a given target material).Case $1$: $V_1 = 3100 \ V$.$\lambda_{min, 1} = \frac{12400}{3100} = 4 \ \mathring{A}$.Difference $D_1 = \lambda_0 - 4$.Case $2$: $V_2 = 12400 \ V$.$\lambda_{min, 2} = \frac{12400}{12400} = 1 \ \mathring{A}$.Difference $D_2 = \lambda_0 - 1$.According to the problem,$D_2 = 2 	imes D_1$.$\lambda_0 - 1 = 2(\lambda_0 - 4)$.$\lambda_0 - 1 = 2\lambda_0 - 8$.$\lambda_0 = 7 \ \mathring{A}$.Thus,$\lambda_{K_{\alpha}} = 7 \ \mathring{A}$.

The difference between $\lambda_{K_{\alpha}}$ and $\lambda_{min}$ becomes twice if the accelerating potential changes from $3100 \ V$ to $12400 \ V$. Find $\lambda_{K_{\alpha}}$ in $\mathring{A}$. (Given $hc = 12400 \ eV \mathring{A}$)

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