Atomic number of anticathode material in an X | vedclass

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(B) According to Moseley's Law for $K_\alpha$ $X$-rays,the frequency $f$ is given by $\sqrt{f} = a(Z - b)$,where $a = \sqrt{\frac{3}{4} R c}$ and $b =

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$0.76$

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(B) According to Moseley's Law for $K_\alpha$ $X$-rays,the frequency $f$ is given by $\sqrt{f} = a(Z - b)$,where $a = \sqrt{\frac{3}{4} R c}$ and $b = 1$.Substituting the values,$\sqrt{f} = \sqrt{\frac{3}{4} 	imes 1.097 	imes 10^7 	imes 3 	imes 10^8} 	imes (41 - 1)$.$\sqrt{f} \approx 4.97 	imes 10^7 	imes 40 = 1.988 	imes 10^9 	ext{ Hz}^{1/2}$.Squaring both sides,$f \approx 3.95 	imes 10^{18} 	ext{ Hz}$.Using $\lambda = \frac{c}{f}$,where $c = 3 	imes 10^8 	ext{ m/s}$.$\lambda = \frac{3 	imes 10^8}{3.95 	imes 10^{18}} \approx 0.76 	imes 10^{-10} 	ext{ m} = 0.76 \,\mathring A$.

Atomic number of anticathode material in an $X$-ray tube is $41$. Wavelength of $K_\alpha$ $X$-ray produced in the tube is ........ $\mathring A$.

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