A deuteron and an alpha particle are accelerated with the same accelerating potential. Which one of the two has 1) greater value of de-Broglie wavelength, associated with it and 2) less kinetic energy? Explain.

To determine which particle, either a deuteron or an alpha particle, has a greater value of de Broglie wavelength associated with it and which one has less kinetic energy, we need to consider their respective properties and the equations related to de Broglie wavelength and kinetic energy.

Greater de Broglie Wavelength:
The de Broglie wavelength of a particle is given by the equation:
λ = h / p

where λ represents the de Broglie wavelength, h is the Planck's constant, and p is the momentum of the particle. Since both particles are accelerated with the same accelerating potential, we can assume that the potential difference accelerates them to the same kinetic energy.

The momentum of a particle is given by the equation:

p = √(2mK)

where m is the mass of the particle and K is the kinetic energy. Comparing the masses of a deuteron and an alpha particle, we find that the deuteron (a nucleus of deuterium, consisting of a proton and a neutron) has a mass of approximately 2 atomic mass units (u), while the alpha particle (a helium nucleus) has a mass of approximately 4u.

Since momentum is directly proportional to mass, we can conclude that the alpha particle, being heavier, will have a smaller momentum compared to the deuteron when accelerated to the same kinetic energy. Therefore, the deuteron will have a greater de Broglie wavelength associated with it.

Less Kinetic Energy:
Since both particles are accelerated with the same potential, the potential energy will be converted into kinetic energy. However, the kinetic energy of a particle is given by the equation:
K = (1/2)mv^2

where m is the mass of the particle and v is its velocity. Since both particles have the same accelerating potential, they will acquire the same velocity.

As mentioned earlier, the deuteron has a smaller mass compared to the alpha particle. Therefore, when accelerated to the same velocity, the deuteron will have less kinetic energy compared to the alpha particle.

In summary, when accelerated with the same potential, the deuteron will have a greater de Broglie wavelength associated with it and less kinetic energy compared to the alpha particle.