Electrons in an X-ray tube are accelerated by a potential difference of 5.0 kV. If an electron produces one photon on impact with the target, what is the minimum wavelength of the resulting X-rays?
(consider h = 6.626 * 10^-34 Js and e = 1.6 * 10^-19 C)

The correct answer is (C) 0.248 nm.

• By the Duane-Hunt law, the minimum wavelength ${\lambda }_{\min }$$\lambda$_{\min} of continuous X-rays produced when accelerated electrons strike a target is given by:
• ${\lambda }_{\min }=\frac{hc}{eV}$$\lambda$_{\min} = $\frac{hc}{eV}$
• Given $h = 6.626$$\times$ 10^{-34} $\text{ J}$$\cdot$$\text{s}$, $c = 3$$\times$ 10^8 $\text{ m/s}$, $e = 1.6$$\times$ 10^{-19} $\text{ C}$, and $V = 5.0$$\text{ kV}$ = 5000 $\text{ V}$:
• ${\lambda }_{\min }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 5000}=\frac{1.9878\times {10}^{-25}}{8\times {10}^{-16}}\approx 0.248475\times {10}^{-9}\text{ m}=0.248\text{ nm}$$\lambda$_{\min} = $\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 5000}$ = $\frac{1.9878 \times 10^{-25}}{8 \times 10^{-16}}$ $\approx$ 0.248475 $\times$ 10^{-9} $\text{ m}$ = 0.248 $\text{ nm}$