Question

An α–particle (mass = M), having an initial kinetic energy K, approaches ‘head–on’, towards a (for–off) stationary nucleus, having an atomic number z. It comes up to a distance, a₀, of the nucleus before (momentarily) coming to rest, and then turning back. The deBroglie wavelengths, associated with this α–particle, in its initial position (when its kinetic energy is K) equals λ₀. The relation, between λ₀ and a₀, is

Answer 1

Correct option :  (2)

Explanation :

As the [positively charged, α–particle having a charge (+ 2e)] approaches the stationary (positively charged) nucleus (having a charge + ze), its kinetic energy keeps on decreasing and becomes (momentarily) zero when it is at a distance a₀ from the nucleus. At this position, K equals the potential energy of the ‘nucleus alpha particle’ system. We, therefore, have

The deBroglie wavelength, associated with the α–particle, when its kinetic energy is K, is given by 

This is the required relation.