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(B) The shortest wavelength (cut-off wavelength) of $X$-rays is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$.Substituting the values: $h = 6

Vedclass · Accepted Answer

$0.50$

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(B) The shortest wavelength (cut-off wavelength) of $X$-rays is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$.Substituting the values: $h = 6.62 	imes 10^{-34} 	ext{ J-s}$, $c = 3 	imes 10^8 	ext{ m/s}$, $e = 1.6 	imes 10^{-19} 	ext{ C}$, and $V = 25,000 	ext{ V}$.$\lambda_{\min} = \frac{6.62 	imes 10^{-34} 	imes 3 	imes 10^8}{1.6 	imes 10^{-19} 	imes 25,000}$.$\lambda_{\min} = \frac{19.86 	imes 10^{-26}}{40,000 	imes 10^{-19}} = \frac{19.86 	imes 10^{-26}}{4 	imes 10^{-15}} = 4.965 	imes 10^{-11} 	ext{ m}$.Converting to $\mathring{A}$: $4.965 	imes 10^{-11} 	ext{ m} = 0.4965 	imes 10^{-10} 	ext{ m} \approx 0.50 \mathring{A}$.

An $X$-ray machine has an accelerating potential difference of $25,000$ volts. By calculation, the shortest wavelength will be obtained as....... $\mathring{A}$ ($h = 6.62 \times 10^{-34} \text{ J-s}$; $e = 1.6 \times 10^{-19} \text{ C}$).

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