MP Board · Class 12 · Physics · Dual Nature of Radiation and MatterDiscuss de Broglie's hypothesis of matter waves. Derive the expression for the de Broglie wavelength of an electron accelerated through a potential difference of $V$ volts, and calculate its numerical value for $V = 100\text{ V}$.

de Broglie's Hypothesis of Matter Waves\nIn 1924, the French physicist Louis de Broglie proposed a bold hypothesis that nature loves symmetry. Since radiation (like light) exhibits dual character—sometimes behaving like waves and sometimes like particles—matter (like electrons, protons, and atoms) should also exhibit dual behavior.

\nAccording to de Broglie, a wave is associated with every moving particle. These waves are known as matter waves or de Broglie waves. The wavelength associated with a particle of momentum $p$ is given by the de Broglie relation: $$\lambda = \frac{h}{p} = \frac{h}{mv}$$\nwhere:

• $\lambda$ is the de Broglie wavelength,
• $h$ is Planck's constant,
• $p$ is the linear momentum of the particle,
• $m$ is the mass of the particle, and
• $v$ is the velocity of the particle.

Derivation of de Broglie Wavelength for an Accelerated Electron\nConsider an electron of mass $m$ and charge $e$ initially at rest, accelerated from rest through a potential difference of $V$ volts. \nThe work done on the electron by the electric field appears as its kinetic energy ($K$):

$$K = eV$$\nWe know that the kinetic energy and linear momentum $p$ are related by the expression: $$K = \frac{p^2}{2m} \implies p^2 = 2mK = 2meV$$\nTherefore, the momentum $p$ is: $$p = \sqrt{2meV}$|\nSubstituting this value of momentum into the de Broglie wavelength formula $\lambda = \frac{h}{p}$, we get: $$\lambda = \frac{h}{\sqrt{2meV}}$$ \nSubstituting the standard values of constants:

• Planck's constant, $h = 6.63 \times 10^{-34}\text{ J s}$
• Mass of an electron, $m = 9.1 \times 10^{-31}\text{ kg}$
• Charge of an electron, $e = 1.6 \times 10^{-19}\text{ C}$

$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times V}}$$ $$\lambda = \frac{1.227}{\sqrt{V}}\text{ nm} = \frac{12.27}{\sqrt{V}},\text{\AA}$$

Numerical Calculation for $V = 100\text{ V}$\nUsing the derived formula for an electron accelerated through $V = 100\text{ V}$:

$$\lambda = \frac{12.27}{\sqrt{100}},\text{\AA}$$ $$\lambda = \frac{12.27}{10},\text{\AA} = 1.227,\text{\AA} = 0.1227\text{ nm}$$

Conclusion:\nThe de Broglie wavelength of an electron accelerated through $100\text{ V}$ is $1.227,\text{\AA}$.