Question

For a given kinetic energy which of the following has smallest de Broglie wavelength?
(a) Electron
(b) Proton
(c) Deutron
(d) a-particle

Answer 1

(d) a-particle

mark as brillienst answer

Answer 2

Answer:

The alpha particle will have the smallest de Broglie wavelength i.e.option(d).

Explanation:

First, let's see the relation between kinetic energy and de Broglie wavelength,

$\lambda=\frac{h}{\sqrt{2mK}}$          (1)

Where,

λ=wavelength of the light

h=planks constant

m=mass of the particle

K=kinetic energy

As for a given kinetic energy equation (1) can be written as,

$\lambda\propto\frac{1}{\sqrt{m}}$        (2)   (∴ h and K are constant)

From equation (2) we can see that the de Broglie wavelength is inversely proportional to the mass of the particle.

So, let's compare the mass of given particles in option i.e.

$m_ \alpha > m_D > m_P > m_e$       (3)

Where,

$m_\alpha$=mass of the alpha particle

$m_D$=mass of deuteron

$m_P$=mass of the proton

$m_e$=mass of the electron

From equation (2) we know that the wavelength is inversely proportional to mass. Then the order of the wavelength is given as,

$\lambda_e > \lambda_P > \lambda_D > \lambda_\alpha$          (4)

Where,

$\lambda_e$=wavelength of the electron

$\lambda_P$=wavelength of the proton

$\lambda_D$=wavelength of deuteron

$\lambda_\alpha$=wavelength of alpha particle

Hence, after analyzing equation (4) we get to know that the alpha particle will have the smallest de Broglie wavelength.i.e.option(d).