The potential difference between the cathode | vedclass

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(D) The minimum wavelength (cut-off wavelength) of $X$-rays produced in a Coolidge tube is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$.Here

Vedclass · Accepted Answer

$0.123 \ \mathring A$

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(D) The minimum wavelength (cut-off wavelength) of $X$-rays produced in a Coolidge tube is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$.Here,$h = 6.63 	imes 10^{-34} \ J \cdot s$ (Planck's constant),$c = 3 	imes 10^8 \ m/s$ (speed of light),$e = 1.6 	imes 10^{-19} \ C$ (charge of an electron),and $V = 100 \ kV = 10^5 \ V$.Substituting the values:$\lambda_{\min} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{1.6 	imes 10^{-19} 	imes 10^5} \ m$.$\lambda_{\min} = \frac{19.89 	imes 10^{-26}}{1.6 	imes 10^{-14}} \ m$.$\lambda_{\min} \approx 12.43 	imes 10^{-12} \ m = 0.1243 \ \mathring A$.Rounding to the nearest provided option,we get $0.123 \ \mathring A$.

The potential difference between the cathode and the target in a Coolidge tube is $100 \ kV$. The minimum wavelength of the $X$-rays emitted by the tube is

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