# Important Questions for Class 12 Chapter 12 - Atoms

Important questions based on NCERT syllabus for Chapter 12 - Atoms:

Question-1: If you were conducting the alpha particle scattering experiment again, how would you say that the results would vary replacing gold foil by sheets of solid hydrogen knowing that hydrogen solidifies at temperatures below 14k.

Solution:
We know that mass of incident alpha particle (6.64 × 10-27kg) is more than the mass of hydrogen (1.67 × 1027Kg). Hence, the target nucleus is lighter, from which we can infer that alpha particle would not bounce back. Implying to the fact that solid hydrogen isn’t a suitable replacement to gold foil for the alpha particle scattering experiment.

Question-2: The ground state energy of hydrogen atom is −27.2 eV. In this state find the kinetic & potential energies of the electron?

Solution:
Ground state energy of hydrogen atom, E = − 27.2 eV
This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy.
Kinetic energy = − E = − (− 27.2) = 27.2 eV
Potential energy = negative of two times of kinetic energy.
Potential energy = − 2 × ( 27.2 ) = − 54.4 eV

Question-3: Choose a suitable solution to the given statements which justify the difference between Thomson’s model and Rutherford’s model
(a) In the case of scattering of alpha particles by a gold foil, average angle of deflection of alpha particles stated by Rutherford’s model is( less than, almost the same as, much greater than )stated by Thomson’s model.
(b) Is the likelihood of reverse scattering (i.e., dispersing of α-particles at points more prominent than 90°) anticipated by Thomson’s model ( considerably less, about the same, or much more prominent ) than that anticipated by Rutherford’s model?
(c) For a small thickness T, keeping other factors constant, it has been found that amount of alpha particles scattered at direct angles is proportional to T. This linear dependence implies?
(d) To calculate average angle of scattering of alpha particles by thin gold foil, which model states its wrong to skip multiple scattering?

Solution:
(a) almost the same
The normal point of diversion of alpha particles by a thin gold film anticipated by Thomson’s model is about the same as from anticipated by Rutherford’s model. This is on the grounds that the average angle was taken in both models.
(b) much less
The likelihood of scattering of alpha particles at points more than 90° anticipated by Thomson’s model is considerably less than that anticipated by Rutherford’s model.
(c) Dispersing is predominantly because of single collisions. The odds of a single collision increment linearly with the amount of target molecules. Since the quantity of target particles increment with an expansion in thickness, the impact likelihood depends straightly on the thickness of the objective.
(d) Thomson’s model
It isn’t right to disregard multiple scattering in Thomson’s model for figuring out the average angle of scattering of alpha particles by a thin gold film. This is on the grounds that a solitary collision causes almost no deflection in this model. Subsequently, the watched normal scattering edge can be clarified just by considering multiple scattering.