Based on which experiment did the Rutherford nuclear model come from? 

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Ernst Rutherford was engaged in experiments on $\alpha$-particles emitted by some radioactive elements and explanation of the results gave an explanation of the atomic model.

According to this the entire positive charge and most of the mass of the atom is concentrated in small volume called the nucleus with electron revolving around the nucleus just as planets revolve around the Sun which is also called planetary model of atom or Rutherford nuclear model which we have accepted today.

However, it could not explain why atoms emit light of only discrete wavelengths. For example, how could an atom as simple as hydrogen, consisting of a single electron and a single proton, emit a complex spectrum of specific wavelengths?

Similar Questions

Answer the following questions, which help you understand the difference between Thomson's model and Rutherford's model better.

$(a)$ Is the average angle of deflection of $\alpha$ -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 model?

$(b)$ Is the probability of backward scattering (i.e., scattering of $\alpha$ -particles at angles greater than $90^{\circ}$ ) 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 small thickness $t,$ the number of $\alpha$ -particles scattered at moderate angles is proportional to $t$. What clue does this linear dependence on $t$ provide?

$(d)$ In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of $\alpha$ -particles by a thin foil?

In gold foil experiment number of deflected $\alpha -$ particles at angle $90^o$ is $63$ than number of $\alpha -$ particle deflected at $120^o$ is

The energy of hydrogen atom in $n^{th}$ orbit is $E_n$, then the energy in $n^{th}$ orbit of singly ionised helium atom will be

An $\alpha$-particle of energy $4\  MeV$ is scattered through $180^o $ by a fixed uranium nucleus. The distance of the closest approach is of the order of

For principal quantum number $n = 3$, the possible values of orbital quantum number $‘l’$ are