(a) The size of the atom in Thomson’s model is ……. The atomic size in Rutherford’s modal (much greater than/no different from/much less than)
(b) IN the ground state of………., electrons are in stable equilibrium, while in …….. electrons always experience a net force (Thomson’s model/Rutherford’s modal).
(c) A classical atom based on……..is doomed to collapse (Thomson’s model/ Rutherford’s model ).
(d) An atom has a nearly continuous mass distribution in……..but has a highly non uniform mass distribution in………(Thomson’s model/Rutherford’s model).
(e) The positively charged part of the atom possesses most of the mass of atom in….. (Rutherford’s model/both the models).
(b) Thomson’s model, Rutherford’s model.
(c) Rutherford’s model.
(d) Thomson’s model, Rutherford’s model.
(e) Both the models.
Sol. In ground state, energy of gaseous hydrogen at room temperature = -13.6 eV. When it is bombarded with 12.5 eV electron beam, the energy becomes -13.6 + 12.5= -1.1 eV. The electron would
excitation the electron may jump from n = 3 to n = 2 giving rise to Balmer series. It may also jump form n = 3 to n = 1, giving rise to Lyman series.
(which help you understand the difference between Thomson’s model and Rutherford’s model better). (a) Is the average angle of deflection of α particles by a thin gold foil predicted by Thomson’s model much less, about the same or much greater than that predicted by Rutherford’s atom model ?
(b)Is the probability of back ward scattering (i.e. scattering of particles at angles greater than 900) predicted by Thomson’s model, much less, about the same, or much greater than that predicted by Rutherford’s model ?
(c) Keeping other factors fixed, it is found experimentally, that for a small thickness t, the number of alpha particles scattered at moderate angles is proportional to t. What clues does this linear dependence on t provide ?
(d) In which atom model, is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of alpha particles by a thin foil ?
(b)Much less, because in Thomson’s model, there is no such massive central core called the nucleus as in Rutherford’s model.
(c)This suggests that scattering is predominantly due to a single collision, because chance of a single collision increases with the number of target atoms which increases linearly with the thickness of the foil.
(d) In Thomson model, positive charge is uniformly distributed in the spherical atom. Therefore, a single collision causes very little deflection. Therefore, average scattering angle can be explained only by considering multiple scattering. Thus it is wrong to ignore multiple scattering in Thomson’s model.
On the contrary, in Rutherford’s model, most of the scattering comes form a single collision. Therefore, multiply scattering may be ignored as a first approximation.
This much greater than the estimated size of the whole universe !
In Bohr’s atom model,
Hence for large values of n, classical frequency of revolution of electron in nth orbit is the same as the frequency of radiation emitted when hydrogen atom de-excites from level (n) to level (n – 1).
(a) Construct a quantity with the dimensions of length from the fundamental constants e, m e and c. Determine its numerical value.
(b) You will find that the length obtained (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for something else to get the right atomic size. Construct a quantity with the dimensions of length from h, m e and e and confirm that its numerical value has indeed the correct order of magnitude.
this is much smaller than the typical atomic size.
(b) However, when we drop c and use h, m_e and e to construct a quantity, which has dimension of length, the quantity we obtain is
(a) What is kinetic energy of electron in this state ?
(b) What is potential energy of electron in this state ?
(c) Which of the answers above would change if the choice of zero of potential energy is changed ?
P.E. = -2 (Kinetic energy)
In this calculation, electric potential and hence potential energy is zero at infinity.
Total energy = P.E. + K.E. = -2 KE + KE = -KE
(a) In the first excited state, total energy = -3.4 eV ؞ KE = – (-3.4 eV) = + 3.4 eV
(b) P.E. of electron in this first excited state = – 2 KE = – 2 × 3.4 = 3.4 = – 6.8 eV
(c) If zero of potential energy is changed, K. E. does not change and continues to be + 3.4 eV. However, the P.E and total energy of the state would change with the choice of zero of potential energy.