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If the value of Rydberg constant is 1.09...

If the value of Rydberg constant is `1.097xx10^(7) m^(-1)` , what will be the wavelengths of the emitted radiations in case of the following transitions in a hydrogen atom ?
(i) From n=5 to n=2

Text Solution

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The correct Answer is:
(i) 4341 Å
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If the value of Rydberg constant is 1.097xx10^(7) m^(-1) , what will be the wavelengths of the emitted radiations in case of the following transitions in a hydrogen atom ? (ii) From n=5 to n=1

If the value of Rydberg constant is 1.097xx10^(7) m^(-1) , what will be the wavelengths of the emitted radiations in case of the following transitions in a hydrogen atom ? (iii) From n=5 to n=3

Knowledge Check

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

    A
    `0.5xx10^(7)m^(-1)`
    B
    `0.25xx10^7m^(-1)`
    C
    `2.5xx10^7m^(-1)`
    D
    `0.025xx10^4m^(-1)`

  • H, He^(+), Li^(2+) are examples of atoms or ions with one electron each . The energy of such atoms when in the n-th energy state (according to Bohr,s theory , n=1,2,3…. =principal quantum number ) is E_n =(-13.6 Z^2)/(n^2) eV (1 eV =1.6xx10^(-19)J) . For the ground state ,n=1 . in order to raise the atom from the ground state to n=f , the suitable incident light should have a wavelength given by lambda=(hc)/(E_f-E_1) . But the atom cannot stay permanently in the f-energy state, ultimately , it comes to the ground state by radiating the extra energy , E_f-E_1 as electromagnetic radiation . The electron of the atom comes from n=f to n=1 in one or more steps using the permitted energy levels . As a result there is a possibility of emission of radiation with more than one wavelength from the atom. Planck's constant =6.63 xx10^(-34)J*s and velocity of light c=3xx10^(8)m*s^(-1) . The wavelength of radiation emitted for the transition of the electron of He^+ ion from n=4 to n=2 is

    A
    952 Å
    B
    975 Å
    C
    1027 Å
    D
    1219 Å

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