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Q.

Electrons with de-brogile wavelength $λ$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is

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a

$λ_{0} = \frac{2 m c λ^{2}}{h}$

b

$λ_{0} = \frac{2 h}{m c}$

c

$λ_{0} = \frac{2 m^{2} c^{2} λ^{2}}{h^{2}}$

d

$λ_{0} = λ$

answer is A.

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Detailed Solution

$λ = \frac{h}{p} = \frac{h}{\sqrt{2 m k}} \Rightarrow k = \frac{h^{2}}{2 m λ^{2}}$ But, $λ_{0} = \frac{h c}{k}$

From equation (i) and equation (ii) $λ_{0} = \frac{2 m c λ^{2}}{h}$

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