Question

De Broglie wavelength of electrons before hitting a target in an X-ray tube of voltage 2000 volts is (Take mass of electron as $9\times {10}^{-31}$ kg )

• A. 27.5 $pm$
• B. $15A^{\circ }$
• C. 13.6 $nm$
• D. 12.4 $\mu m$

Answer 1

The correct option is A 27.5 $pm$

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mKE}}$ where $KE=e{V}_0$
So $\lambda =\frac{h}{\sqrt{2me{V}_o}}=\frac{6.6\times {10}^{-34}}{\sqrt{2\times 9\times {10}^{-31}\times 1.6\times {10}^{-19}\times 2\times {10}^3}}=\frac{6.6}{24}\times {10}^{-10}m=\frac{660}{24}\times {10}^{-12}m=27.5pm$