The X-rays produced in a Coolidge tube of pot | vedclass

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(A) The minimum wavelength (cutoff wavelength) of $X-$rays produced by an electron accelerated through a potential difference $V$ is given by the form

Vedclass · Accepted Answer

$3.09 	imes 10^{-11} \ m$

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(A) The minimum wavelength (cutoff wavelength) of $X-$rays produced by an electron accelerated through a potential difference $V$ is given by the formula: $\lambda_{\min} = \frac{hc}{eV} = \frac{12400 \ \mathring{A} \cdot V}{V 	ext{ (in volts)}}$.Given $V = 40 \ kV = 40,000 \ V$.Substituting the values: $\lambda_{\min} = \frac{12400}{40000} \ \mathring{A} = 0.31 \ \mathring{A}$.Converting to meters: $0.31 	imes 10^{-10} \ m = 3.1 	imes 10^{-11} \ m$.Given the options provided,the closest value is $3.09 	imes 10^{-11} \ m$.

The $X-$rays produced in a Coolidge tube of potential difference $40 \ kV$ have a minimum wavelength of:

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