Question

In which of the following situations the heavier of the two particles has smaller de Broglie wavelength? The two particles

• A. move with the same speed
• B. move with the same linear momentum
• C. move with the same kinetic energy
• D. have fallen through the same height

Answer 1

Answer: (A), (C), (D)

The correct options are
A

move with the same speed

C

move with the same kinetic energy

D

have fallen through the same height

The matter-wave wavelength for a mass moving with a speed v(v << c) is given by the relation - $\lambda =\frac{h}{mv}=\frac{h}{p}$

1. Let $m_2>m_1$ , where $m_1$ and $m_2$ are the two masses, $m_2$ being the heavier one. If their speeds are equal, we can claim ${\lambda }_1>{\lambda }_2$ , from the de Broglie relation.

2. If the two particles are moving with speeds $v_1$ and $v_2$ such that,

$m_1{v}_1=m_2{v}_2$

$\Rightarrow p_1=p_2=p$

then their de Broglie wavelengths will be the same too, since $\lambda$ depends just upon the linear momentum , which is the same for both particles.

3. Same kinetic energy? This seems interesting. A mass m having speed v has linear momentum p = mv and kinetic energy $K.E=\left(\frac{m{v}^2}{2}\right)=\left(\frac{m^2{v}^2}{2m}\right)=\left(\frac{p^2}{2m}\right)$ . In our problem, masses $m_1$ and $m_2$ with linear momenta $p_1$ and $p_2$ , have the same kinetic energies. This would mean,

$\frac{p_1^2}{2m_1}=\frac{p_2^2}{2m_2}\Rightarrow \frac{m_2}{m_1}=\frac{p_2^2}{p_1^2}>1(\because m_2>m_1\Rightarrow \frac{m_2}{m_1}>1)\Rightarrow p_2>p_1$

Therefore, from the de Broglie relation, ${\lambda }_1>{\lambda }_2$ (wavelength being inversely proportional to linear momentum)

4. Two different masses falling through the same height will acquire the same velocity at the end of the fall, since they both will undergo the same acceleration. Hence, the linear momentum of the heavier mass will be greater than that of the smaller mass,$m_{2}v>m_{1}v$ or $p_2>p_1$. Therefore, for this case too, ${\lambda }_1>{\lambda }_2$

Hence, options (a), (c), and (d) are correct.