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(A) The minimum wavelength (cut-off wavelength) of $X-rays$ produced is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$.Given,potential differe

Vedclass · Accepted Answer

$0.62$

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(A) The minimum wavelength (cut-off wavelength) of $X-rays$ produced is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$.Given,potential difference $V = 20 \ kV = 20,000 \ V$.Using the relation $\lambda_{\min} (	ext{in } \mathring{A}) = \frac{12400}{V (	ext{in volts})}$.Substituting the value of $V$: $\lambda_{\min} = \frac{12400}{20000} \mathring{A}$.$\lambda_{\min} = \frac{1.24}{2} \mathring{A} = 0.62 \mathring{A}$.

$A$ potential difference of $20 \ kV$ is applied across an $X-ray$ tube. The minimum wavelength of $X-rays$ generated (in $\mathring{A}$) is

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