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Sagot :
The potential difference of 2.7 x [tex]10^{4}[/tex] volts is needed.
We have an X - Ray tube to produce X - Rays.
We have to determine what potential difference to produce X-rays with a minimum wavelength of 45.0 pm ?
What is an X - Ray ?
An X - Ray is an electromagnetic wave of high energy and very short wavelength, which is able to pass through many materials opaque to light.
According to the question, we have -
λ = 45 pm = 4.5 x [tex]10^{-11}[/tex] meter
h = 6.63 × [tex]10^{-34}[/tex]Js
c = 3 x [tex]10^{8}[/tex] m/s
We know -
E = hv
where v is frequency.
E = [tex]$\frac{hc}{\lambda}[/tex]
When accelerated through a potential difference of say V volts, then -
E = E(k) = eV
Therefore -
eV = [tex]$\frac{hc}{\lambda}[/tex]
V = [tex]$\frac{hc}{e\lambda }= \frac{6.63\times 10^{-34} \times 3\times 10^{8} }{1.6\times 10^{-19} \times 4.5 \times 10^{-11} }[/tex] = 2.7 x [tex]10^{4}[/tex] volts
Hence, the potential difference of 2.7 x [tex]10^{4}[/tex] volts is needed.
To solve more questions on X-rays, visit the link below-
https://brainly.com/question/16814407
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