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The cut-off wavelength when a potential difference of 25 kV is applied to an X-ray tube is ____

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$0.248 \overset{\circ}{A}$

$0.496 \overset{\circ}{A}$

$0.124 \overset{\circ}{A}$

$4.96 \overset{\circ}{A}$

answer is B.

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Detailed Solution

$λ_{\min} = \frac{h c}{e V_{p d}} λ_{\min} = \frac{12400 e V A^{o}}{e V p d} λ_{\min} = \frac{12400}{V p d} A^{o} = \frac{12400}{25 \times 10^{3}} A^{o} = 0.496 A^{o}$

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