In a Coolidge tube experiment, the minimum wa | vedclass

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(B) The minimum wavelength of the continuous $X-$ray spectrum is given by the formula $\lambda_{\min} = \frac{hc}{eV}$.Given $\lambda_{\min} = 66.3\,

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electrons accelerate through a potential difference of $18.75\, kV$ in the Coolidge tube.

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(B) The minimum wavelength of the continuous $X-$ray spectrum is given by the formula $\lambda_{\min} = \frac{hc}{eV}$.Given $\lambda_{\min} = 66.3\, pm = 66.3 	imes 10^{-12}\, m$.Using $hc \approx 12400\, eV\cdot\mathring{A} = 1.24 	imes 10^{-6}\, eV\cdot m$, we have:$V = \frac{hc}{e\lambda_{\min}} = \frac{1.24 	imes 10^{-6}}{66.3 	imes 10^{-12}}\, V$.$V = \frac{1.24}{66.3} 	imes 10^6\, V \approx 0.018702 	imes 10^6\, V = 18.702\, kV \approx 18.75\, kV$.Thus, the electrons are accelerated through a potential difference of approximately $18.75\, kV$.

In a Coolidge tube experiment, the minimum wavelength of the continuous $X-$ray spectrum is equal to $66.3\, pm$. Then:

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