Knowing that the minimum x-ray wavelength produced by 40.0 keV electrons striking a target is 31.1 pm, determine the Planck constant .

1. $\mathrm{V}=40\mathrm{keV}$Minimum wavelength produced by the electrons,${\mathrm{\lambda }}_{\min }=31.1\mathrm{pm}$
2. Accelerating potential of the electrons,

Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the magnetic frequency of the photon and thus, equally, equates to the wavelength of the wave. When the frequency of photons is high, its potential is high.

Using Planck's relation and the given values of the minimum wavelength and energy, we can get the value of Planck's constant.

Formula:

The energy of the photon due to Planck’s relation is,

$\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ ….. (1)

Here, the speed of light,

$\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}$

The energy generated due to accelerating potential is,

E = eV ….. (2)

Here the charge is,

$\mathrm{e}=1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}$

As the energy of the emitted electrons is same as the energy due to photon emission, using the given data in combined equations (1) and (2), the value of Planck’s constant isas follows:

$$\begin{gathered} \mathrm{eV}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\min }} \\ \mathrm{h}=\frac{{\mathrm{eV\lambda }}_{\min }}{\mathrm{c}} \end{gathered}$$

Substitute known values in the above equation.

role="math" localid="1661502579390" $$\begin{gathered} \mathrm{h}=\frac{\left(1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)\left(40\times {10}^3\mathrm{eV}\right)\left(31.1\times {10}^{-12}\mathrm{m}\right)}{3\times {10}^8\mathrm{m}/\mathrm{s}} \\ =6.63\times {10}^{-34}\mathrm{J}.\mathrm{s} \end{gathered}$$

Hence, the value of the constant is$6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}$ .