MP Board · Class 12 · Physics · Dual Nature of Radiation and MatterState de Broglie's hypothesis for matter waves. Derive the expression for the de Broglie wavelength of an electron accelerated through a potential difference of $V$ volts. Calculate the de Broglie wavelength associated with an electron accelerated through a potential difference of $100\text{ V}$.

de Broglie's Hypothesis\nIn 1924, the French physicist Louis de Broglie proposed a revolutionary hypothesis that nature loves symmetry. Since radiation (like light) exhibits a dual nature (wave-like and particle-like properties), matter (like electrons, protons, and atoms) should also exhibit wave-like properties under suitable conditions.

\nAccording to de Broglie, a wave is associated with every moving particle of matter. This wave is known as a matter wave or de Broglie wave. The wavelength ($\lambda$) associated with a particle of mass $m$ moving with velocity $v$ is given by the formula: $$\lambda = \frac{h}{p} = \frac{h}{mv}$|\nwhere $h$ is Planck's constant and $p$ is the linear momentum of the particle.

Derivation of de Broglie Wavelength for an Accelerated Electron\nLet an electron of rest mass $m$ and charge $e$ be accelerated from rest through a potential difference of $V$ volts. \nThe kinetic energy ($K$) gained by the electron is given by:

$$K = eV$$\nWe also know that kinetic energy and linear momentum ($p$) are related by the equation: $$K = \frac{p^2}{2m} \implies p^2 = 2mK = 2meV$$ $$p = \sqrt{2meV}$$\nSubstituting the value of momentum $p$ into de Broglie's equation ($\lambda = \frac{h}{p}$), we get: $$\lambda = \frac{h}{\sqrt{2meV}}$$\nThis is the general expression for the de Broglie wavelength of an accelerated charged particle.

Calculation of de Broglie Wavelength for $V = 100\text{ V}$\nSubstitute the known constants into the derived formula:

• $h = 6.63 \times 10^{-34}\text{ J s}$
• $m = 9.1 \times 10^{-31}\text{ kg}$
• $e = 1.6 \times 10^{-19}\text{ C}$
• $V = 100\text{ V}$

$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times (9.1 \times 10^{-31}) \times (1.6 \times 10^{-19}) \times 100}}$$ $$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{291.2 \times 10^{-50}}} = \frac{6.63 \times 10^{-34}}{17.065 \times 10^{-25}}$$ $$\lambda = 0.3885 \times 10^{-9}\text{ m} = 0.388\text{ nm} = 1.227\text{ \u00c5}$$

Final Answer:

• de Broglie Wavelength expression: $\lambda = \frac{h}{\sqrt{2meV}}$
• Calculated wavelength for $100\text{ V}$: $1.227\text{ \u00c5}$ (or $0.1227\text{ nm}$)