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The electronic transition from n=2 to n=...

The electronic transition from n=2 to n=1 will produce shortest wavelength in (where n=principal quantum state)-

A

`Li^(2+)`

B

`He^(2+)`

C

`H`

D

`H^(+)`

Text Solution

Verified by Experts

The correct Answer is:
A
`(1)/(lamda)=Z^(2)*R_(H)[(1)/(1^(2))-(1)/(2^(2))]=(3)/(4)R_(H)Z^(2)" Therefore, "lamdaprop(1)/(Z^(2))`
So, higher the value of Z, shorter will be the value of wavelength, thus `Li^(2+)` will have shorter wavelength.
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(1) Show that the sum of energies for the transition from n=3 to n=2 and from n=2 to n=1 is equal to the energy of transition from n=3 to n=1 in case of an H-atom. (2) Are wavelength and frequencies of the emitted spectrum also additive as their energies?

The electron in a hydrogen atom makes a transition n_1 to n_2 , where n_1 and n_2 are the principal quantum numbers of the two states. According to the Bohr model , the time period of revolution of electron in initial state is 8 times that in the final state. Which are possible values of n_1 and n_2 ? A 4,2 , B 8,2 , C 8,1 ,D 6,3 .

Knowledge Check

  • The electronic transition from n=2 to n=1 will produce shortest wavelength in (where n= principal quantum number)

    A
    `Li^(2+)`
    B
    `He^+`
    C
    H
    D
    `H^+`

  • Electron in a hydrogen atom undergoes transition from n_1 to n_2 , where n_1 and n_2 are the principal quantum numbers of two given states. According to Bohr's theory , if the time period of the electron in the initial state be eight times that in the final state , the possible values of n_1 and n_2 will be, respectively

    A
    4 and 2
    B
    8 and 2
    C
    8 and 1
    D
    6 and 4

  • 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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