Question

Which of the following has the largest de-Broglie wavelength, given that all have equal velocity?
A. $C{O_2}$ molecule
B. $N{H_3}$ molecule
C. Electron
D. Proton

Answer 1

The de-Broglie equation says that the wavelength is equal to Planck's constant divided by mass multiplied by velocity. As velocity is the same for all then the wavelength is dependent on the mass.

Complete step by step answer:
It is given that the given molecules have equal velocity.
The de-Broglie explained that the matter can show wave-particle duality. The wavelength is given by the equation shown below.
$\lambda = \dfrac{h}{p}$
Where,
$\lambda$ is the wavelength
h is the planck's constant
p is momentum
When the particle is moving with velocity, the momentum is given as shown below.
$p = mv$
Where,
P is momentum
m is mass
v is velocity.
Substitute the value of momentum in the equation, so the de-Broglie wavelength is given by the equation as shown below.
$\lambda = \dfrac{h}{{mv}}$
-As velocity is the same for all the compounds and particles, then the wavelength is only dependent on the mass.
$\lambda \alpha \dfrac{1}{m}$
-The wavelength is inversely proportional to the mass.
As the mass of the compound increases the wavelength decreases.
-The mass of carbon dioxide is 44 g.
-The mass of ammonia is 17 g.
-The mass of the photon is $1.672 \times {10^{ - 27}}Kg$.
-The mass of the electron is $9.109 \times {10^{ - 31}}Kg$.
-The mass of the electron is less as compared to other compounds and particles , thus its wavelength will be maximum.
Therefore, the correct option is C.

Note:
In the de-Broglie equation, the mass m is the relative mass and not the rest of the mass as the rest of the mass of the photon is zero.