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When a particle is restricted to move along the x-axis between x = 0 and x = a, where 'a' is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass, 'm' is related to its linear momentum as E=p2/2m. Thus, the energy of the particle can be denoted by a quantum number 'n' taking values 1,2,3,.... (n=1, called the ground state) corresponding to the number of loops in the standing wave. 
Take h=6.6×1034Jsande=1.6×1019C.

The speed of the particle that a particle can take in discrete values is proportional to ( 'n' represents 'nth' orbit) 
  1. n
  2. n3/2
  3. n1/2
  4. n1


Solution

The correct option is A n
mv=nh2πr or v=nh2πrm  v α n

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