An electron is accelerated at 10 kV to a tungsten target. The wavelength of X-ray photon produced will be

0.62 A

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2.48 A

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0.124 A

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0.032 A

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Solution

The correct option is D 0.032 A
$V = 10, 000 V$
$∴ E n e r g y, E = e V = 1.6 \times 10^{- 19} \times 10, 000 J$
$λ = \frac{h}{\sqrt{2 m E}}$
$= \frac{6.64 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 10^{4} \times 1.6 \times 10^{- 19}}}$
$= 0.032 \times 10^{- 10} m = 0.032 A$

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An electron is accelerated at 10 kV to a tungsten target. The wavelength of X-ray photon produced will be

The correct option is D 0.032 AV=10,000V∴Energy,E=eV=1.6×10−19×10,000Jλ=h√2mE=6.64×10−34√2×9.1×10−31×104×1.6×10−19=0.032×10−10m=0.032A

The correct option is D 0.032 A
$V = 10, 000 V$
$∴ E n e r g y, E = e V = 1.6 \times 10^{- 19} \times 10, 000 J$
$λ = \frac{h}{\sqrt{2 m E}}$
$= \frac{6.64 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 10^{4} \times 1.6 \times 10^{- 19}}}$
$= 0.032 \times 10^{- 10} m = 0.032 A$