An X-ray tube is working at potential of 20 kV. The potential difference is decreased to 10 kV. It is found that the difference of the wavelength of $K_{α}$ X-ray and the most energetic continuous X-ray becomes 4 times the difference before the change of voltage. The atomic number of the target element is _______________ Take b=1 and $\frac{1}{\sqrt{3.4}} = 0.54$

Moderate

Solution

$\begin{array}{l} We use λ_{min} = \frac{h c}{20000 e}, λ_{min}^{'} = \frac{h C}{10000 e} \\ Given that 4 (λ_{α} - λ_{min}) = (λ_{α} - λ_{min}^{'}) \\ \Rightarrow 3 λ α = 4 λ_{min} - λ_{min}^{'} \\ \Rightarrow λ_{α} = \frac{h C}{30000 e} \Rightarrow \frac{h C}{λ_{α}} = 30000 e \end{array}$
By using the formula of energy of electrons according to Bohr's model and considering shielding effect
$\begin{array}{l} 30000 = (13.6) (\frac{1}{12} - \frac{1}{2^{2}}) (Z - 1)^{2} \\ \Rightarrow Z - 1 = 100 \sqrt{\frac{5}{17}} = \frac{100}{\sqrt{34}} = 54 \\ Z = 54 + 1 = 55 \end{array}$

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