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If R is the Rydberg's constant for hydro...

If `R` is the Rydberg's constant for hydrogen the wave number of the first line in the Lyman series will be

A

`(R )/(4)`

B

`(3R)/(4)`

C

`(R )/(2)`

D

`2R`

Text Solution

Verified by Experts

The correct Answer is:
B
For Lyman series
`bar(v) = (1)/(lambda) = R ((1)/(1^(2)) - (1)/(n^(2)))` here `n = 2, 3, 4, 5` …….
For first line
`bar(v) = R ((1)/(1^(2)) - (1)/(2^(2))) rArr bar(v) = R (1-(1)/(4)) = (3R)/(4)`
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Using the Rydberg formula, calculate the wavelength of the first four spectral lines in the Lyman series of the hydrogen spectrum.

Knowledge Check

  • The Rydberg constant R for hydrogen is

    A
    `R = -((1)/(4pi epsilon_(0))).(2pi^(2)me^(2))/(ch^(2))`
    B
    `R = ((1)/(4pi epsilon_(0))).(2pi^(2)me^(4))/(ch^(2))`
    C
    `R =((1)/(4pi epsilon_(0)))^(2).(2pi^(2)me^(4))/(c^(2)h^(2))`
    D
    `R = ((1)/(4pi epsilon_(0)))^(2).(2pi^(2)me^(4))/(ch^(3))`

  • Rydberg's constant for hydrogen line is-

    A
    `109678 cm^(-1)`
    B
    `109.678 cm^(-1)`
    C
    `1096.78 cm^(-1)`
    D
    `10.9678 cm^(-1)`

  • The wave number of the series limit of Lyman series is -

    A
    `1.097 xx 10^(7) m^(-1)`
    B
    `2.74 xx 10^(6) m^(-1)`
    C
    `1.22 xx 10^(6) m^(-1)`
    D
    `6.86 xx 10^(5) m^(-1)`

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    In terms of Rydberg's constant R , the wave number of the first Balman line is

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    Given the value of Rydberg constant is 10^(7)m^(-1) , the waves number of the lest line of the Balmer series in hydrogen spectrum will be:

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