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

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.