Question

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

Answer 1

Hint: Recollect the basic concepts of quantum mechanics. The de Broglie equation is given as, $\lambda =\dfrac{h}{mv}$ where $\lambda $ is the wavelength, h is the Planck’s constant, m is the molecular weight and v is the velocity of the molecules. Just substitute the values while keeping in mind units of each parameter to get the answer.

Complete step by step answer:
- Let’s take a look at the question and write down the given data.
- Velocity is given so, v = $2.4\times {{10}^{5}}cm\,{{s}^{-1}}$.
- Molecular weight of hydrogen molecule is 2g/mol.
- de Broglie equation tells us about the wave nature of a particle such as an electron.
- de Broglie equation gives the relationship between the mass and velocity of a particle and the wavelength. According to de Broglie equation, wavelength of a particle is inversely proportional to its mass and velocity.
- de Broglie equation is given as, $\lambda =\dfrac{h}{mv}$ where $\lambda $ is the wavelength, h is the Planck’s constant, m is the molecular weight and v is the velocity of the molecule.
- We know, Planck’s constant h = $6.6\times {{10}^{-27}}erg/s$
- Therefore, substituting the values in the de Broglie equation we get,
\[\lambda =\dfrac{h}{{}^{m}/{}_{{{N}_{a}}}\times v}=\dfrac{6.6\times {{10}^{-27}}erg/s}{{}^{2}/{}_{6.023\times {{10}^{23}}}g\times 2.4\times {{10}^{5}}cm.{{s}^{-1}}}\]
\[\therefore \lambda =8.28\times {{10}^{-9}}cm\]
- So, we obtained de Broglie wavelength as $8.28\times {{10}^{-9}}cm$ but the options are given in an angstrom unit.
-Therefore, de Broglie wavelength,
$\Rightarrow \lambda =8.28\times {{10}^{-9}}cm=0.828\times {{10}^{-8}}cm=0.83A{}^\circ \approx 1A{}^\circ $
- Hence, the de Broglie wavelength in this case is approximately equal to $1A{}^\circ $.

So, the correct answer is “Option A”.

Note: Remember de Broglie equation tells us about the dual nature of light to act as a particle as well as a wave. de Broglie equation gives the relation between wavelength of light which is inversely proportional to the mass and velocity of a particle. Remember one angstrom unit is ${{10}^{-10}}m$ or ${{10}^{-8}}cm$.