Answer to Question #120308 in Physics for daniel

Question

Q4. If you wish to produce 45.0 nm x-rays in the laboratory, what is the minimum voltage you must use in accelerating the electrons?

Answer 1

In our case the minimum voltage is the photon energy with the wavelength of 45.0 nm:

$$E_{photon} = \dfrac{hc}{\lambda},$$ $$E_{photon} = \dfrac{6.626 \cdot 10^{-34} \ J \cdot s \cdot 3 \cdot 10^8 \ \dfrac{m}{s}}{45 \ nm},$$ $$E_{photon} = \dfrac{1.9878 \cdot 10^{-25} \ J \cdot m \cdot \dfrac{1 \ eV}{1.602 \cdot 10^{-19} \ J}}{45 \ nm},$$ $$E_{photon} = \dfrac{1240 \cdot 10^{-9} \ eV \cdot m \cdot \dfrac{10^9 \ nm}{m}}{45 \ nm},$$ $$E_{photon} = \dfrac{1240 \ eV \cdot nm}{45 \ nm} = 27.5 \ eV.$$

Answer:

$E_{photon} = 27.5 \ eV.$