If the short wavelength limit of the continuous spectrum coming out of a Coolidge tube is 10 $\overset{o}{A}$ , then the de Broglie wavelength of the electrons reaching the target metal in the Coolidge tube is approximately

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10 $\overset{o}{A}$

30 $\overset{o}{A}$

3 $\overset{o}{A}$

0.3 $\overset{o}{A}$

answer is A.

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Detailed Solution

$\begin{array}{l} We have KE = \frac{P^{2}}{2 m_{e}} = \frac{h c}{λ_{min}} \\ P = \sqrt{\frac{2 h {cm}_{e}}{λ_{min}}} \\ Also, λ_{de Broglie} = \frac{h}{p} = \sqrt{\frac{h λ_{min}}{2 m_{e} c}} \\ For λ_{min} = 10 {Å, c=3×10}^{8} {m/s , h=6.6×10}^{- 34} {Js, m}_{e} {=9.1×10}^{- 31} kg \\ \begin{matrix} w e g e t λ_{de Broglie} ≅ 0.3 Å \end{matrix} \end{array}$