The following particles are moving with the same velocity. The particle with maximum de Broglie wavelength is
(A) Electron
(B) Proton
(C) Neutron
(D) $ \alpha - {\text{particles}} $
Answer
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Hint : The de Broglie wavelength is inversely proportional to the momentum of a body. Momentum of massive bodies is given as the product of their masses and velocities.
Formula used: In this solution we will be using the following formulae;
$ \lambda = \dfrac{h}{p} $ where $ \lambda $ is the de Broglie wavelength of a particle, $ h $ is the Planck’s constant, $ p $ is the momentum of the body. $ p = mv $ where $ p $ is the momentum of a massive particle (i.e. possesses mass), $ m $ is the mass of the particle, and $ v $ is the velocity of the particle at the moment.
Complete step by step answer:
A list of particles are given to us, being said that they possess the same velocity, the particle with the largest de Broglie wavelength is to be determined.
Now, the de Broglie formula is usually written as $ \lambda = \dfrac{h}{p} $ where $ \lambda $ is the de Broglie wavelength of a particle, $ h $ is the Planck’s constant, $ p $ is the momentum of the body. And of course, momentum (for massive particles) is given by
$ p = mv $ where $ p $ is the momentum of a massive particle (i.e. possesses mass), $ m $ is the mass of the particle, and $ v $ is the velocity of the particle at the moment.
Hence, $ \lambda = \dfrac{h}{{mv}} $
This implies, that for the same velocity of all the particles, the higher the mass, the lower the wavelength, hence, the particle with the lower mass has the highest wavelength.
From observation of the options, we can be well certain that electrons has the lowest mass amongst them. Hence, the particle with the maximum wavelength is the electron.
Hence the correct option is A.
Note:
For clarity, we should note the emphasis placed on the momentum of the massive particle. This is because only the momentums of massive particles have their momentum defined as such. Massless particles, such as photons cannot be defined as $ p = mv $ but by some other formula. Massless particles do have momentum.
Formula used: In this solution we will be using the following formulae;
$ \lambda = \dfrac{h}{p} $ where $ \lambda $ is the de Broglie wavelength of a particle, $ h $ is the Planck’s constant, $ p $ is the momentum of the body. $ p = mv $ where $ p $ is the momentum of a massive particle (i.e. possesses mass), $ m $ is the mass of the particle, and $ v $ is the velocity of the particle at the moment.
Complete step by step answer:
A list of particles are given to us, being said that they possess the same velocity, the particle with the largest de Broglie wavelength is to be determined.
Now, the de Broglie formula is usually written as $ \lambda = \dfrac{h}{p} $ where $ \lambda $ is the de Broglie wavelength of a particle, $ h $ is the Planck’s constant, $ p $ is the momentum of the body. And of course, momentum (for massive particles) is given by
$ p = mv $ where $ p $ is the momentum of a massive particle (i.e. possesses mass), $ m $ is the mass of the particle, and $ v $ is the velocity of the particle at the moment.
Hence, $ \lambda = \dfrac{h}{{mv}} $
This implies, that for the same velocity of all the particles, the higher the mass, the lower the wavelength, hence, the particle with the lower mass has the highest wavelength.
From observation of the options, we can be well certain that electrons has the lowest mass amongst them. Hence, the particle with the maximum wavelength is the electron.
Hence the correct option is A.
Note:
For clarity, we should note the emphasis placed on the momentum of the massive particle. This is because only the momentums of massive particles have their momentum defined as such. Massless particles, such as photons cannot be defined as $ p = mv $ but by some other formula. Massless particles do have momentum.
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