91Ó°ÊÓ

The wavelength of the x-ray, $\mathrm{λ}=0.1nm$

One electron-volt kinetic energy is acquired by an electron or proton working at a potential change of one volt. In terms of cost and potential difference, the cost formula is

E =eV

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.

Formulas:

The kinetic energy gained by the electron is,

$\mathrm{Δ}E=eV$ ….. (1)

Where, e is the charge and V is the accelerating potential difference.

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

$E=hf$

$E=\frac{hc}{\mathrm{λ}}$ ….. (2)

Here, h is the Plank’s constant, c is the speed of light, f is the frequency, and $\mathrm{λ}$is the wavelength.

Consider the known data as below.

The Plank’s constant, $h=6.63×{10}^{-34}Jâ‹\dots s$

The speed of light,$c=3×{10}^8\frac{m}{s}$

The charge,$e=1.6×{10}^{-19}\frac{J}{eV}$

The wavelength,$\mathrm{λ}=0.100nm=0.1×{10}^{-3}m$

Equating both the equations (1) and (2), the value of the minimum potential difference is given by,

$$\begin{gathered} eV=\frac{hc}{{\mathrm{λ}}_{m}ax} \\ V=\frac{hc}{\mathrm{λ}e} \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} V=\frac{\left(6.63×{10}^{-34}J.s\right)\left(3×{10}^{8}m/s\right)}{\left(0.1×{10}^{-3}m\right)\left(1.6×{10}^{-19}J/eV\right)} \\ =1.24×{10}^{4}eV \\ =12.4kV \end{gathered}$$

Hence, the value of the potential difference is 12.4 kV.