When p.d applied across an x-ray tube is reduced by 23 kV the minimum wavelength of the continuous x-rays spectrum emitted by this tube is doubled. What was the original value of the minimum wavelength ?

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0.27 Å

1.1 Å

0.13 Å

0.54 Å

answer is B.

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Detailed Solution

$λ_{min} = \frac{12400}{V_{pd}} \overset{0}{A} \Rightarrow λ_{min} \propto \frac{1}{V_{pd}}$

$\Rightarrow \frac{λ_{1}}{λ_{2}} = \frac{V_{2}}{V_{1}}$

$\Rightarrow \frac{λ}{2 λ} = \frac{V - 23}{V} \Rightarrow V_{pd} = 46 kV$

$\Rightarrow λ_{min} = \frac{12400}{V_{pd}} \overset{0}{A} = \frac{12400}{46 \times 10^{3}} \overset{0}{A} = 0.27 \overset{0}{A}$

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