Question

consider a proton, an electron, and a helium nucleus (consisting of 2 protons and 2 neutrons) moving at the same speed vv. list their de broglie wavelengths from largest to smallest.

Answer 1

The de Broglie wavelengths from largest to smallest are: electron, proton, helium nucleus.

Step-by-step explanation:

The de Broglie wavelength of a particle is given by the equation:

λ = h / p

where λ is the de Broglie wavelength, h is the Planck's constant, and p is the momentum of the particle. The momentum of a particle can be calculated using the equation:

p = mv

where m is the mass of the particle and v is its velocity. In this case, we are comparing the de Broglie wavelengths of a proton, an electron, and a helium nucleus moving at the same speed.

Let's compare their de Broglie wavelengths:

1. The de Broglie wavelength of a proton is given by:
2. λ = h / (mv)
3. where m is the mass of the proton and v is its velocity. Since the proton and the helium nucleus have the same speed, their de Broglie wavelengths will be the same.
4. The de Broglie wavelength of an electron is given by the same equation as above. The mass of an electron is much smaller than that of a proton or a helium nucleus, so its de Broglie wavelength will be larger.
5. Since the de Broglie wavelength is inversely proportional to the mass, the helium nucleus will have the smallest de Broglie wavelength.

Therefore, the de Broglie wavelengths from largest to smallest are: electron, proton, helium nucleus.