Question

The mass of a moving particle is m and velocity is v, then write down the formula for de-Broglie wavelength $\lambda $ .

Answer 1

Hint: The De-Broglie equation is derived from Einstein's energy equation. Energy is the product of Planck’s constant and frequency which is equated with the product of mass and square of the speed. Since frequency is the ratio of velocity to the wavelength, it is substituted in place of v. On rearranging the terms, we get the equation for wavelength $\lambda = \dfrac{h}{{mv}}$ .

Complete Step by step solution
De Broglie began deriving the wavelength equation by a series of substitutions.
Starting with Einstein’s energy mass relation: $E = m{c^2}$
Where, E is the energy of a particle of mass m moving with speed of light c.
Subsequently taking the energy expression given by Planck’s equation: $E = h\nu $
Where, E is the discrete energy of a quantum wave packet having frequency, v.
Equating the two expressions of energy,
 $ \Rightarrow E = m{c^2} = h\nu $
For real particles, the speed with which they travel cannot be equal to the speed of light, c. So, we will replace the speed ‘c’ in the above equation by speed ‘v’ at which our particle/ wave packet is travelling.
 $ \Rightarrow E = m{v^2} = h\nu $
Now using, $v = \nu \lambda $ or $\nu = \dfrac{v}{\lambda }$ in the above equation,
 $
   \Rightarrow m{v^2} = h\dfrac{v}{\lambda } \\
   \Rightarrow \lambda = h\dfrac{v}{{m{v^2}}} \\
   \Rightarrow \lambda = \dfrac{h}{{mv}} \\
 $

Note: We can also use the direct formula for De Broglie’s wavelength $\lambda = \dfrac{h}{p}$ where p is momentum of the particle and simply substitute $p = mv$ to obtain the same answer.