Question

Two particles A₁ sand A₂ of masses m₁, m₂ (m₁ > m₂) have the same de Broglie wavelength. Then ______.

1. their momenta are the same.
2. their energies are the same.
3. energy of A₁ is less than the energy of A₂.
4. energy of A₁ is more than the energy of A₂.

विकल्प

• a and c

• a and d

• c and d

• a and b

Answer 1

a and c

Explanation:

We know that de-Broglie wavelength `λ = h/(mv)`

Where, mv = p (momentum) of the particle

⇒ But we can express wavelength `λ = h/p` ⇒ `p = h/λ`

Hence, `p  oo  1/λ ⇒ p_1/p_2 = λ_2/λ_1`

But particles have the same de-Broglie wavelength `(λ_1 = λ_2) = λ`

Then `p_1/p_2 = λ/λ` = 1 ⇒ `p_1 = p_2`

Thus, their momenta are the same.

Also, `E = 1/2  mv^2 = 1/2 mv^2 m/m`

= `1/2 (m^2v^2)/m = 1/2 p^2/m`

As p is constant, `E  oo 1/m`

∴ `E_1/E_2 = m_2/m_1 < 1 ⇒ E_1 < E_2`

Important points: 

Some important characteristics of Matter Waves:

1. Matter wave represents the probability of finding a particle in space.
2. Matter waves are not electromagnetic in nature.
3. de-Broglie or matter wave is independent of the charge on the material particle. It means matter-wave of de-Broglie wave is associated with every moving particle (whether charged or uncharged).
4. Practical observation of matter waves is possible only when the de-Broglie wavelength is of the order of the size of the particles.
5. Electron microscope works on the basis of de-Broglie waves.
6. The phase velocity of the matter waves can be greater than the speed of light.
7. Matter waves can propagate in a vacuum, hence they are not mechanical waves.
8. The number of de-Broglie waves associated with n^th orbital electron is n.
9. Only those circular orbits around the nucleus are stable whose circumference is an integral multiple of de-Broglie wavelength associated with the orbital electron.