Write the expression for the Broglie wavelength associated with a charged particle having charge 'q' and mass 'm', when it is accelerated by a potential V.

Open in App

Solution

When the charge particle is accelerated by a potential V, the kinetic energy is equal to electrostatic potential energy.
Thus, $\frac{1}{2} m v^{2} = q V$ or $v = \sqrt{2 q V / m}$
Hence, the de Broglie wavelength, $λ = \frac{h}{p} = \frac{h}{m v} = \frac{h}{\sqrt{2 m q V}}$

Suggest Corrections

Similar questions

Q. Write the expression for the de-Broglie wavelength associated with a charged particle having charge

q

and mass

m

, when it is accelerated by a potential

V

A charged particle of mass ‘m’ and charge ‘q’ is accelerated through a potential difference ‘V’. The De Broglie wavelength associated with the particle is

Q. A proton when accelerated through a potential difference of

V

, has a de Broglie wavelength

λ

associated with it. If an alpha particle is to have the same de Broglie wavelength

λ

, it must be accelerated through a potential difference of:

Q. A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of

50 V

. Another particle B of mass '4 m' and charge 'q' is accelerated by a potential difference of

2500 V

The ratio of de-Broglie wavelengths

\frac{λ_{A}}{λ_{B}}

is close to :

Derive an expression for de Broglie wavelength associated with an electron accelerated through a potential v.

Join BYJU'S Learning Program

Write the expression for the Broglie wavelength associated with a charged particle having charge 'q' and mass 'm', when it is accelerated by a potential V.

When the charge particle is accelerated by a potential V, the kinetic energy is equal to electrostatic potential energy. Thus, 12mv2=qV or v=√2qV/mHence, the de Broglie wavelength, λ=hp=hmv=h√2mqV

When the charge particle is accelerated by a potential V, the kinetic energy is equal to electrostatic potential energy.
Thus, $\frac{1}{2} m v^{2} = q V$ or $v = \sqrt{2 q V / m}$
Hence, the de Broglie wavelength, $λ = \frac{h}{p} = \frac{h}{m v} = \frac{h}{\sqrt{2 m q V}}$