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(A) Balmer series lies in the visible re...

(A) Balmer series lies in the visible region of electromagnetic spectrum
(R): `(1)/(lambda) = R((1)/(2^(2)) - (1)/(n^(2)))` where n = 3,4,5

A

2651 Å

B

1641.1 Å

C

6569 Å

D

3249 Å

Text Solution

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The correct Answer is:
B
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n_(1) value in Balmer series is

In case of hydrogen spectrum wave number is given by barv = R_H [(1)/(n_2^2) - 1/(n_2^2)] where n_1 lt n_2

Knowledge Check

  • The observed wavelengths in the line spectrum of hydrogen atom were first expressed in terms of a series by Johann Jakob Balmer, a Swiss teacher. Balmer's empirical formula is (1)/(lambda) = R_(H)[(1)/(2^(2)) - (1)/(n^(2))] = n = 3, 4,5… R_(H) = 109678 cm^(-1) is the Rydberg constant. Niels Bohr derived this expression theoretically in 1913. The formula is a generalised to any on electron atom/ion. The wavelength of first line of Balmer spectrum of hydrogen will be :

    A
    4340 Å
    B
    4101 Å
    C
    6569 Å
    D
    4861 Å

  • The observed wavelengths in the line spectrum of hydrogen atom were first expressed in terms of a series by Johann Jakob Balmer, a Swiss teacher. Balmer's empirical formula is (1)/(lambda) = R_(H)[(1)/(2^(2)) - (1)/(n^(2))] = n = 3, 4,5… R_(H) = 109678 cm^(-1) is the Rydberg constant. Niels Bohr derived this expression theoretically in 1913. The formula is a generalised to any on electron atom/ion. Which of the following is not correctly matched ?

    A
    `H_(alpha)` - 6569 Å (Red)
    B
    `H_(beta)` - 4861 Å (Blue)
    C
    `H_(gamma)` - 4340 Å (Orange)
    D
    `H_(delta)` -4101 Å (Violet)

  • The observed wavelengths in the line spectrum of hydrogen atom were first expressed in terms of a series by Johann Jakob Balmer, a Swiss teacher. Balmer's empirical formula is (1)/(lambda) = R_(H)[(1)/(2^(2)) - (1)/(n^(2))] = n = 3, 4,5… R_(H) = 109678 cm^(-1) is the Rydberg constant. Niels Bohr derived this expression theoretically in 1913. The formula is a generalised to any on electron atom/ion. How many lines in the spectrum will be observed when electrons returns from 7th shell to 2nd shell ?

    A
    13
    B
    14
    C
    15
    D
    16

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    Lt_(n rarr oo) sum_(r=1)^(n)[(1)/(sqrt(4n^(2) - r^(2)))]

    The observed wavelengths in the line spectrum of hydrogen atom were first expressed in terms of a seriesby Johamn Balmer a swiss teacher. Balmer.s empirical formula is : (1)/(lambda) = R_(H)[(1)/(2^(2))-(1)/(n^(2))]n=3,4,5 …. R_(H) = 109678 cm^(-1) is the Rydberg constant The wavelength of first line of Balmer spectrum of hydrogen will be :

    The observed wavelengths in the line spectrum of hydrogen atom were first expressed in terms of a seriesby Johamn Balmer a swiss teacher. Balmer.s empirical formula is : (1)/(lambda) = R_(H)[(1)/(2^(2))-(1)/(n^(2))]n=3,4,5 …. R_(H) = 109678 cm^(-1) is the Rydberg constant How many lines in the spectrum will be observed when electrons return from 7^(th) shell to 2^(nd) shell ?

    Balmer gave an equation for wavelength of visible region of H-spectrum as barv =(n^2 - 4)/(Kn^2) where n = principal quantum number of energy level, K = constant terms.of R (Rydberg constant). The value of K in terms of R is:

    When electron jumps from higher orbit to lower orbit, then energy is radiated in the form of electro magnetic radiation and these radiations are used to record the emission spectrum Energy of electron may be calculated as E= (2pi^2 m_e Z^2 e^4)/(n^2 h^2) where m_e = rest mass of electron DeltaE = (E_(n_2)- E_(n_1)) = 13.6xxZ^2 xx [1/(n_1^2) - 1/(n_2^2)] eV per atom This equation was also used by Rydberg to calculate the wave number of a particular line in the spectrum. bar(upsilon) = 1/(lambda) = R_H Z^2 [1/(n_1^2) -1/n_2^2] m^(-1) Where , R_H = 1.1 xx 10^7 m^(-1) (Rydberg constant) For Lyman, Balmer, Paschen, Brackett and Pfund series the value of n_1 = 1,2,3,4,5 respectively and n_2 = oo for series limit, If an electron jumps from higher orbit n to ground state than number of spectral line will be ""^nC_2 . Ritz modified the Rydberg equation by replacing the rest mass of electron with reduced mass (mu) . 1/(mu) = 1/(m_N) + 1/(m_e) , Here m_N mass of nucleus , m_e = mass of electron. The emission spectrum of He^+ involves transition of electron from n_1 to n_2 such that n_2 + n_1 = 8 and n_2 - n_1 = 4 . What will be the total number of lines in the spectrum ?

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