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mario62 [17]
3 years ago
5

The results of Rutherford's gold foil experiment gave him the evidence to arrive at two conclusions: (1) an atom was much more t

han just empty space and scattered electrons and (2) ______________________________________________.
Chemistry
1 answer:
Vedmedyk [2.9K]3 years ago
7 0

Answer: The results of Rutherford's gold foil experiment gave him the evidence to arrive at two conclusions: (1) an atom was much more than just empty space and scattered electrons and (2) an atom consists of a positive charge at the center where most of its mass is placed.

Explanation:

Ernest Rutherford performed an experiment in which he passed alpha particles through a thin gold foil sheet. Through this sheet some of the particles passed on to the other side but some of them were reflected back.

This experiment was done by Rutherford to prove than an atom contains a tiny and heavy nucleus. He concluded that an atom has more than empty space and electrons were present in scattered form.

Also, he concluded that an atom consists of a positive charge at the center where most of its mass is placed.

Thus, we can conclude that the results of Rutherford's gold foil experiment gave him the evidence to arrive at two conclusions: (1) an atom was much more than just empty space and scattered electrons and (2) an atom consists of a positive charge at the center where most of its mass is placed.

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The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

                                   \frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} ),

where  λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from n_1 to n_2 and R_h = 109677 is the Rydberg Constant. Here n_1 and  n_2 represents the transitions.

(a) n_1 =2 to n_2 = infinity

            \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2}  ) = 109677/4     [since 1/infinity = 0] Therefore, \lambda = 4 / 109677 = 0.00003647 m

(b) n_1=4 to  n_2 = 20

           \frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2}  ) = 6580.62

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(c) n_1=3 to  n_2 = 10

          \frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2}  ) = 11089.56

Therefore,  \lambda = 1 / 11089.56 = 0.00009 m

(d)  n_1=2 to  n_2 = 1

          \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2}  ) = - 82257.75

Therefore,  \lambda = 1 /82257.75  = - 0.0000121 m  

[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]

So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.

Learn more about the Rydberg's formula athttps://brainly.com/question/14649374

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