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(A) The minimum wavelength $(\lambda_{min})$ of $X$-rays produced by electrons accelerated through a potential difference $V$ is given by the Duane-Hu

Vedclass · Accepted Answer

$0.62$

Vedclass · Answer

(A) The minimum wavelength $(\lambda_{min})$ of $X$-rays produced by electrons accelerated through a potential difference $V$ is given by the Duane-Hunt law: $\lambda_{min} = \frac{hc}{eV}$.Substituting the values: $h = 6.63 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$,$e = 1.6 	imes 10^{-19} \ C$,and $V = 20 	imes 10^3 \ V$.$\lambda_{min} = \frac{(6.63 	imes 10^{-34}) 	imes (3 	imes 10^8)}{(1.6 	imes 10^{-19}) 	imes (20 	imes 10^3)} \ m$.$\lambda_{min} = \frac{19.89 	imes 10^{-26}}{3.2 	imes 10^{-15}} \ m = 6.215 	imes 10^{-11} \ m$.Since $1 \ Å = 10^{-10} \ m$,we have $\lambda_{min} = 0.6215 	imes 10^{-10} \ m = 0.62 \ Å$.

The minimum wavelength of $X$-rays produced by $20 \ kV$ electrons is nearly (in $Å$)

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