Question

calculate the atomic number of the element which gives minimum X ray wavelength of 0.1 nm of k series R=1.09737×10^7
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Answer 1

Answer:

0.041nm

Explanation:

Calculate the minimum wavelength of X-rays emitted when electrons accelerated through 30 kV strike a target.

f = [1.6 x 10-19 x 3 x 104]/ 6.63 x 10-34 = 7.2x1018 Hz

Therefore the wavelength l (= c/f) is 0.41 x 10-10 m = 0.04 1 nm (compared with some 600 nm for yellow light).

Answer 2

The atomic number $Z$ ≅ 36

Explanation:

Given :

Wavelength of X-ray $\lambda$ = $0.1 \times 10^{-9}$ m

Rydburg constant $R = 1.09737 \times 10^{7}$

From the moseley's law,

⇒  $\sqrt{f} = k (Z-1)$

We have wavelength of X-ray, to find frequency of X-ray.

   $c = f \lambda$

   $f = \frac{3 \times 10^{8} }{0.1 \times 10^{-9} } = 30 \times 10^{17}$Hz

Where $f =$ frequency of X-ray, $k = 4.965 \times 10^{7}$, $Z =$ atomic number.

        $f = k^{2} (Z-1)^{2}$

$(Z-1)^{2} = \frac{30 \times 10^{17} }{24.65 \times 10^{14} }$

   $Z-1 = 34.88$

        $Z$ ≅ 36