Question

At ${{200}^{o}}$ C, hydrogen molecule have velocity \[2.4\times {{10}^{5}}cm{{s}^{-1}}\]. The de Broglie wavelength in this case is approximately:
A.1 $A{}^\circ $
B.1000 $A{}^\circ $
C.100 $A{}^\circ $
D.10 $A{}^\circ $

Answer 1

Hint: Wavelength that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength. A particles de Broglie wavelength which is shown by the symbol λ is inversely proportional to its force.

Complete answer:
It is said that matter has a dual nature of wave particles. But it is actually a property of a material object that varies in time or space while behaving similar to waves these type of waves are also called matter waves. De Broglie wavelength is given by
\[\lambda =\dfrac{h}{p}\]
Where h is planck’s constant \[6.6\times {{10}^{-27}}\]ergs
p is the linear momentum which is actually given by the product of mass and volume i.e. \[p=m\times v\]
Hence According to question
\[\lambda =\dfrac{h}{mv}\]
v = \[2.4\times {{10}^{5}}cm{{s}^{-1}}\]
Mass of one molecule of hydrogen = \[\dfrac{2}{{{N}_{a}}}\]; where \[{{N}_{a}}\]represents the avogadro’s number having value \[6.022\times {{10}^{23}}\].
Mass of hydrogen = \[\dfrac{2}{6.022\times {{10}^{23}}}\]
Now put all the values in
\[\lambda =\dfrac{h}{mv}\]
\[=\dfrac{6.6\times {{10}^{-27}}\times 6.022\times {{10}^{23}}}{2\times 2.4\times {{10}^{5}}}\]
\[8.27\times {{10}^{-9}}cm\]
\[=0.8\times {{10}^{-8}}cm\]
= 0.8 $A{}^\circ $
Thus, we can say that approximately wavelength is equal to 1 $A{}^\circ $;

So option A is correct.

Note:
De Broglie wavelength has another type called thermal de Broglie wavelength which is approximately the average de Broglie wavelength of the gas particles in an ideal gas at the specified temperature.