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

(a) move with the same speed
(c) move with the same kinetic energy
(d) have fallen through the same height

Let m₁ be the mass of the heavier particle and m₂ be the mass of the lighter particle.
If both the particles are moving with the same speed v, de Broglie wavelength of the heavier particle,
${\lambda }_1=\frac{h}{m_{1}v}$ ...(1)
de Broglie wavelength of the lighter particle,
${\lambda }_2=\frac{h}{m_{2}v}$ ...(2)
Thus, from equations (1) and (2), we find that if the particles are moving with the same speed v, then ${\lambda }_1<{\lambda }_2$.
Hence, option (a) is correct.

If they are moving with the same linear momentum, then using the de Broglie relation
$\lambda =\frac{h}{p}$
We find that both the bodies will have the same wavelength. Hence, option (b) is incorrect.

If K is the kinetic energy of both the particles, then de Broglie wavelength of the heavier particle,
${\lambda }_1=\frac{h}{\sqrt{2m_{1}K}}$
de Broglie wavelength of the lighter particle,
${\lambda }_2=\frac{h}{\sqrt{2m_{2}K}}$
It is clear from the above equation that if $m_1>m_2$, then ${\lambda }_1<{\lambda }_2$.
Hence, option (c) is correct.

When they have fallen through the same height h, then velocity of both the bodies,
v = $\sqrt{2gh}$
Now,
${\lambda }_1=\frac{h}{m_1\sqrt{2gh}}$
${\lambda }_2=\frac{h}{m_2\sqrt{2gh}}$
m₁>m₂
$\therefore$ ${\lambda }_1<{\lambda }_2$
Hence, option (d) is correct.