Question
According to de-Broglie, a moving material particle sometimes acts as a wave and sometimes as a particle or a wave associated with moving material particle which controls the particle in every respect. The wave associated with moving particle is called matter wave or de-Broglie wave where wavelength called de-Broglie wavelength, rs given by $\lambda=\frac{\text{h}}{\text{mv}}$.
  1. If a proton and an electron have the same de Broglie wavelength, then:
  1. Kinetic energy of electron < kinetic energy of proton.
  2. Kinetic energy of electron = kinetic energy of proton.
  3. Momentum of electron = momentum of proton.
  4. momentum of electron < momentum of proton.
  1. Which of these particles having the same kinetic energy has the largest de Broglie wavelength?
  1. Electron
  2. Alpha particle
  3. Proton
  4. Neutron
  1. Two particles $A_1$ and $A_2$ of masses $m_1, m_2 (m_1 > m_2)$ have the same de Broglie wavelength. Then:
  1. Their momenta are the same.
  2. Their energies are the same.
  3. Momentum of $A_1$ is less than the momentum of $A_1$.
  4. Energy of $A_1$ is more than the energy of $A_2.$
  1. When the velocity of an electron increases, its de Broglie wavelength:
  1. Increases.
  2. Decreases.
  3. Remains same.
  4. May increase or decrease.
  1. Proton and $\alpha$ - particle have the same de-Broglie wavelength. What is same for both of them?
  1. Time period
  2. Energy
  3. Frequency
  4. Momentum

Answer

  1. (c) Momentum of electron = momentum of proton.
Explanation:
de Broglie wavelength, $\lambda=\frac{\text{h}}{\text{p}}$
where pis the momentum of the particle
For electron, $\lambda_\text{e}=\frac{\text{h}}{\text{p}_\text{e}}$
For proton, $\lambda_\text{p}=\frac{\text{h}}{\text{p}_\text{p}}$
As $\lambda_\text{e}=\lambda_\text{p}\Rightarrow\text{P}_\text{e}=\text{P}_\text{p}$ (Given) or
Momentum of electron = Momentum of proton
  1. (a) Electron
Explanation:
As $\lambda=\frac{\text{h}}{\sqrt{2\text{mk}}}$ so $\lambda\propto\frac{1}{\sqrt{\text{m}}}$
Out of the given particles rn is least for electron, therefore electron has the largest value of de Broglie wavelength.
  1. (a) Their momenta are the same.
Explanation:
As $\lambda=\frac{\text{h}}{\text{p}}$ or $\text{p}=\frac{\text{h}}{\lambda}$ or $\text{p}\propto\frac{1}{\lambda}$
$\therefore\frac{\text{p}_1}{\text{p}_2}=\frac{\lambda_2}{\lambda_1}=\frac{\lambda}{\lambda}=1$ or $\text{p}_1=\text{p}_2$
Also $\text{E}=\frac{1}{2}\frac{\text{p}^2}{\text{m}}=\frac{1}{2\text{m}}\frac{\text{h}^2}{\lambda^2}$ $\Big(\because\text{p}=\frac{\text{h}}{\lambda}\Big)$
or $\text{E}\propto\frac{1}{\text{m}}\therefore\frac{\text{E}_1}{\text{E}_2}=\frac{\text{m}_2}{\text{m}_1}<1 \text{ or E}_1<\text{E}_2$
  1. (b) Decreases.
Explanation:
The de Broglie wavelength is given by
$\lambda=\frac{\text{h}}{\text{p}}=\frac{\text{h}}{\text{mv}}$
So if the velocity of the electron increases, the de Broglie wavelength decreases.
  1. (d) Momentum
Explanation:
$\lambda=\frac{\text{h}}{\text{p}},$ when $\lambda$ is same, p is also same.

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