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The ratio of lowest energy in ierms of w...

The ratio of lowest energy in ierms of wave number of balmer and lyman series of lines of atomic specturm of hydrogen is

A

`5:27`

B

`27:5`

C

`20:27`

D

`27:2`

Text Solution

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The correct Answer is:
A
` :.` wave number ` (v) = R Z ^(2) [ (1)/(n _(L^(2)))-(1)/(n_(H ^(2)))]`
wave number of lowest energy for balmer series `(n_(L) = 2 ,n_(H) =3)`
`v = R [ (1) /(4) - (1) /(9)] = (5) /(36) R`
wave number for lowest energy for lyman series :
`( n_(L) = 2 , n_(H) = 2 ) implies " " v = R [1 -(1) /(4)] = (3)/(4) R`
Thus , ratio of Balmer / lyman is
`= (R5 //36)/(R3//4)= (5xx4)/(3xx36) = (5)/(27)`
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The only electron in the hydrogen atom resides under ordinary conditions on the first orbit. When energy is supplied, the electron moves to higher energy orbit depending on the amount of energy absorbed. When this electron returns to any of the lower orbits, it emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electron returns to the third, fourth orbits from higher energy orbits respectively (as shown in figure) Maximum number of lines produced when an electron jumps from nth level to ground level is equal to (n(n-1))/(2) . For example, in the case of n = 4, number of lines produced is 6. (4 rarr 3, 4 rarr 2, 4 rarr 1, 3 rarr 2, 3 rarr 1, 2 rarr 1) . When an electron returns from n_(2) to n_(1) state, the number of lines in the spectrum will be equal to ((n_(2) - n_(1))(n_(2)-n_(1) +1))/(2) If the electron comes back from energy level having energy E_(2) to energy level having energy E_(2) then the difference may be expressed in terms of energy of photon as E_(2) - E_(1) = Delta E, lambda = (h c)/(Delta E) . Since h and c are constant, Delta E corresponds to definite energy, thus each transition from one energy level to another will prouce a higher of definite wavelength. THis is actually observed as a line in the spectrum of hydrogen atom. Wave number of the line is given by the formula bar(v) = RZ^(2)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) Where R is a Rydberg constant (R = 1.1 xx 10^(7)) (i) First line of a series : it is called .line of logest wavelength. or .line of shortest energy.. (ii) Series limit of last of a series : It is the line of shortest wavelength or line of highest energy. The difference in the wavelength of the 2^(nd) line of Lyman series and last line of Bracket series in a hydrogen sample is