Matter Wave

1 Mark Questions

1. 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 [All India 2014] Ans.

A charged particle having charge q and mass m, then

kinetic energy of the particle is equal to the work done on it by the electric field.

i.e. <math>K = qV</math> <math>\Rightarrow</math> <math display="block">\frac{1}{2}mv^2 = qV</math> <math>\Rightarrow</math> <math display="block">\frac{p^2}{2m} = qV</math> ⇒ <math>p = \sqrt{2mqV}</math> (1) 2.State de-Broglie hypothesis. [Delhi 2012]

Ans.

de-Broglie hypothesis A Moving object sometimes acts as a wave and sometimes as a particle or a wave is associated with the moving particle which control this particle in every respect. This wave associated with the moving particle is called matter wave or de-Broglie wave. Its wavelength is given by

<math>v = \frac{h}{}</math> mν

where, <math>h = Planck</math> constant, <math>m = mass</math> of object, <math>v = velocity</math> of the object. (1)

3.A proton and an electron have same kinetic energy. Which one has greater de-Broglie wavelength and why? [All India 2012]

Ans.

de-Broglie wavelength,

<math display="block">\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}</math>

where, <math>K = KE</math>

For given KE,

<math>\Rightarrow \lambda \propto \frac{1}{\sqrt{m}}</math>

Electron have smaller mass, <math>\lambda_e > \lambda_p</math>

<math>[: m_e < m_p]</math>

For given kinetic energy, electrons have greater wavelength as these have smaller mass. (1)

1. Write the relationship of de-Broglie wavelength X associated with a particle of mass m in terms of its kinetic energy E. [Delhi 2011c]

Ans.

Kinetic energy, <math>K = \frac{p^2}{2m}</math> <math>p = momentum</math> ٠, <math>m = mass</math> and <math>K = \text{kinetic energy}</math>

<math>\Rightarrow p = \sqrt{2mK}</math>

de-Broglie wavelength, <math>\lambda = \frac{n}{p}</math>

where, <math>p = \sqrt{2mK}</math><br> <math display="block">\Rightarrow \lambda = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mE}} [K = E] (1)</math>

1. Show graphically, the variation of de-Broglie wavelength (A,) with the potential (V) through

which an electron is accelerated from rest. [Delhi 2011]

Ans.