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Question

A potential difference of 20kV is applied across an X-ray tube. The minimum wavelength of X-rays generated is ____________ A.


A

1.621A°

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B

0.621A°

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C

2.621A°

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D

5.621A°

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Solution

The correct option is B

0.621A°


Explanation for correct option

Step 1: Given data

Potential difference V=20kV

=20000V

e=1.6×10-19Columbus

c=3×108m/s

Step 2: Find the wavelength

The amount of energy obtained by a charged particle when it is accelerated through a potential V is,

E=eV ----------- 1

Here, e is charge, and V is potential.

The wavelength of the emitted X-ray is,

λ=hceV ---------- 2

Here, h is planks constant, λ is wavelength, e is charge and c is velocity of light.

Substitute the given values in equation 2. We get,

λ=6.626×10-34×3×1081.6×10-19×20000

=1.98×10-253.2×10-15

=0.62×10-10m

=0.62A°

Therefore, the wavelength of x-ray generated is 0.62A°

Hence, option B is correction answer.


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