Question

The wavelength associated with a gold weighing 200g and moving at a speed of 5 \[m/h\] is of the order.
 A. \[{10^{ - 10}}\] m
 B. \[{10^{ - 20}}\] m
 C. \[{10^{ - 30}}\] m
 D. \[{10^{ - 40}}\] m

Answer 1

Hint: The distance between two adjacent crests of a wave is called wave length. Wavelength of a wave can be measured in meters, centimeters, or millimeters units of length. The human eye can detect the wavelengths from about 380 to 700 nanometers. In this range the highest wavelength for red light and lowest is for violet.
Formula used:
  \[\lambda = \dfrac{h}{{mv}}\] , where m and v is mass and velocity of the particle respectively.

Complete step by step answer:
 Wavelength of any wave depends upon the energy of that wave. Higher the energy of the wave lower will be the wavelength and vice-versa. Wavelength is inversely proportional to the frequency of the wave. i.e. with increasing frequency wavelength decreases and vice-versa. Wavelength of a wave can be related with the mass and velocity of the particle moving in that wave. According to the de-Broglie all particles have waves like nature.
The formula to calculate wavelength is,
  \[E = \dfrac{{hc}}{\lambda }\] where wave length is \[\lambda \] ,velocity of light is c, the Planck constant is h \[(6.626 \times {10^{ - 34}}{m^2}kg/\sec )\] .
And, according to de Broglie \[\lambda = \dfrac{h}{{mv}}\] , where m and v is mass and velocity of the particle respectively.
Now from given data,
  \[
  v = 5m/hr = \dfrac{5}{{60 \times 60}}m/\sec \\
  m = 200gm = 200 \times {10^{ - 3}}kg \\
\]
The wavelength is,
\[
  \lambda = \dfrac{h}{{mv}} \\
\Rightarrow \lambda = \dfrac{{6.626 \times {{10}^{ - 34}} \times 60 \times 60}}{{200 \times {{10}^{ - 3}} \times 5}} \\
\Rightarrow \lambda = 2.3853 \times {10^{ - 30}}m \\
 \]
This value of wavelength is the order of \[{10^{ - 30}}\] m .

So, the correct option is C.

Note:
The de-Broglie states that an electron has both particle nature and wave nature. This is also called the dual nature of electrons. It can be said that electron beams can be diffracted like a beam of light. The formula of de-Broglie wavelength is the ratio of Planck constant and momentum of a particle, where mv is the momentum of the particle.