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Which of the following is cprrect regard...

Which of the following is cprrect regarding Bohr's theory?
(i) Elecrtronic energy is quantized.
(ii) Bohr showed the physical meaning of the two whole numbers `n` and `m` in the Balmer-Rydeberg equation.
(iii) Electrons can only be in certain discrete orbits.
(iv) Electrons absorb or emit enegry in discrete amounts as they more form one orbit to anothe.

A

`(i),(ii),(iii),(iv)`

B

`(i),(ii),(iii)`

C

`(ii),(iii),(iv)`

D

`(i),(ii),(iii),(iv)`

Text Solution

Verified by Experts

The correct Answer is:
D

(i) Only certain values of electronic enegry are possible.
(ii) The whole numbers `n` and `m` represent the two enegry states between which the transition takes place.
(iii) The radius `(r )` of the circular orbit proportional to `n^(2)`.
(iv) Bohr's application of the idea of quantization of energy to the electron in an atom provided the answer to the puzzle concerning the discrete colors given off in the spectrum.
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One of the fundamental laws of physics is that matter is most stable with the lowest possible energy. Thus, the electron in a hydrogen atom usually moves in the n=1 orbit, the orbit in which it has the lowest energy. When the electon is in this lowest energy orbit, the atom is said to be in its ground electronic state. If the atom receives energy from an outside source, it is possible for the electron to move ot an orbit with a higher n value, in which case the atoms is in an excited state with a higher energy. The law of conservation of energy says that we cannot create or destroy energy. Thus, if a certain amount of external energy is required to excite an electron from one energy level to another, then that same amount of energy will be liberated when the electron returns to its initial state. Lyman series is observed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electrons returns to the third, fourth and fifth orbits from higher energy orbits respectively. When electrons return form n_(2) " to " n_(1) state, the number of lines in the spectrum will equal to ((n_(2)-n_(1))(n_(2)-n_(1)+1))/(2) If the electon comes back from energy level having energy E_(2) to energy level having energy E_(1) , then the difference may be expressed in terms of energy of photon as : E_(2)-E_(1)=DeltaE, deltaE implies (hc)/(lambda) Since, h and c are constant, deltaE corresponds to definite energy. Thus, each transition from one energy level to another will produce a radiatiob of definite wavelength. This is actually Wave number of a spectral line is given by the formula barv=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2))) . where R is a Rydberg's constant (R=1.1xx10^(7) m^(-1)) What transition in the hydrogen spectrum would have the same wavelength as Balmer transitio, n=4 " to "n=2 in the He^(+) spectrum?

One of the fundamental laws of physics is that matter is most stable with the lowest possible energy. Thus, the electron in a hydrogen atom usually moves in the n=1 orbit, the orbit in which it has the lowest energy. When the electon is in this lowest energy orbit, the atom is said to be in its ground electronic state. If the atom receives energy from an outside source, it is possible for the electron to move ot an orbit with a higher n value, in which case the atoms is in an excited state with a higher energy. The law of conservation of energy says that we cannot create or destroy energy. Thus, if a certain amount of external energy is required to excite an electron from one energy level to another, then that same amount of energy will be liberated when the electron returns to its initial state. Lyman series is observed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electrons returns to the third, fourth and fifth orbits from higher energy orbits respectively. When electrons return form n_(2) " to " n_(1) state, the number of lines in the spectrum will equal to ((n_(2)-n_(1))(n_(2)-n_(1)+1))/(2) If the electon comes back from energy level having energy E_(2) to energy level having energy E_(1) , then the difference may be expressed in terms of energy of photon as : E_(2)-E_(1)=DeltaE, deltaE implies (hc)/(lambda) Since, h and c are constant, deltaE corresponds to definite energy. Thus, each transition from one energy level to another will produce a radiatiob of definite wavelength. This is actually Wave number of a spectral line is given by the formula barv=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2))) . where R is a Rydberg's constant (R=1.1xx10^(7) m^(-1)) If the wavelength of series limit of Lyman series for He^(+) ion is x Å, then what will be the wavelength of series limit of Balmer series for Li^(2+) ion?

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