The minimum wavelength of X-ray emission by an X-ray tube is $0.4125 A^{o}$ The accelerating voltage is :

30 k V

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50 k V

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80 k V

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60 k V

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Solution

The correct option is D $30 k V$
From the formula,
$e v = h \frac{c}{λ}$
$v = \frac{h c}{e λ}$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19} \times 0.4125 \times 10^{- 10}}$
$= 30 \times 10^{3} V = 30 k V$

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The minimum wavelength of X-ray emission by an X-ray tube is0.4125Ao The accelerating voltage is :

The correct option is D 30kVFrom the formula, ev=hcλv=hceλ=6.6×10−34×3×1081.6×10−19×0.4125×10−10=30×103V=30kV

The minimum wavelength of X-ray emission by an X-ray tube is $0.4125 A^{o}$ The accelerating voltage is :

The correct option is D $30 k V$
From the formula,
$e v = h \frac{c}{λ}$
$v = \frac{h c}{e λ}$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19} \times 0.4125 \times 10^{- 10}}$
$= 30 \times 10^{3} V = 30 k V$