To produce characteristic X-rays using a tungsten target in an X-ray generator, the energy of the characteristic radiation is $eV .$
(The binding energy of the innermost electron in tungsten is $40 keV .)$

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Solution

For characteristic X-ray, electron should jump from
$n = 2$ to $n = 1$
the wavelength of X-ray is given by
$\frac{1}{λ} = R_{α} (Z - 1)^{2} [1 - \frac{1}{n^{2}}] = \frac{E}{h c}$
$\frac{E}{h c} = R (Z - 1)^{2} (1 - \frac{1}{2^{2}}) . . . (i)$
Binding energy $E_{b} = 40 keV$
$\frac{E_{b}}{h c} = R (Z - 1)^{2} (1 - \frac{1}{\infty^{2}}) . . . (i i)$

$\frac{E}{E_{b}} = \frac{[1 - \frac{1}{2^{2}}]}{[1 - \frac{1}{\infty^{2}}]}$
$\Rightarrow \frac{E}{E_{b}} = [1 - \frac{1}{4}]$
$E = \frac{3}{4} E_{b} = \frac{3}{4} \times 40, 000 = 30, 000 eV$

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To produce characteristic X-rays using a tungsten target in an X-ray generator, the energy of the characteristic radiation is eV. (The binding energy of the innermost electron in tungsten is 40 keV.)

For characteristic X-ray, electron should jump from n=2 to n=1 the wavelength of X-ray is given by 1λ=Rα(Z−1)2[1−1n2]=Ehc Ehc=R(Z−1)2(1−122)...(i) Binding energy Eb=40 keV Ebhc=R(Z−1)2(1−1∞2)...(ii) EEb=[1−122][1−1∞2] ⇒EEb=[1−14] E=34Eb=34×40,000=30,000 eV

To produce characteristic X-rays using a tungsten target in an X-ray generator, the energy of the characteristic radiation is $eV .$
(The binding energy of the innermost electron in tungsten is $40 keV .)$