An electron beam in X-ray tube is accelerated through a potential difference of 50,000volt. These are then made to fall on a tungsten target. The shortest wavelength of the X-rays emitted by the tube is

2.5 A^{0}

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0.25 n m

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0.25 c m

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0.025 n m

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Solution

The correct option is D $0.025 n m$
$λ_{c u t o f} f = \frac{h c}{e v}$

$= \frac{20 \times 10^{- 26}}{1.6 \times 10^{- 19} \times 5 \times 10^{4}}$

$= 0.025 n m$
So, the answer is option (D).

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An electron beam in X-ray tube is accelerated through a potential difference of 50,000volt. These are then made to fall on a tungsten target. The shortest wavelength of the X-rays emitted by the tube is

The correct option is D 0.025nmλcutoff=hcev =20×10−261.6×10−19×5×104 =0.025nmSo, the answer is option (D).

The correct option is D $0.025 n m$
$λ_{c u t o f} f = \frac{h c}{e v}$

$= \frac{20 \times 10^{- 26}}{1.6 \times 10^{- 19} \times 5 \times 10^{4}}$

$= 0.025 n m$
So, the answer is option (D).