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The frequency of radiation absorbed or e...

The frequency of radiation absorbed or emitted when transition occurs between two stationary states with energies `E_(1)` (lower) and `E_(2)` (higher) is given by

A

`v=(E_(1)+E_(2))/h`

B

`v=(E_(1)-E_(2))/h`

C

`v=(E_(1)xxE_(2))/h`

D

`v=(E_(2)-E_(1))/h`

Text Solution

Verified by Experts

The correct Answer is:
D
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Using Bohr's postualtes , derive the expression for the frequency of radiation emitted when electron in hydrogen atom undergoes transition from higher energy state ( quantum number n_(i) ) to the lower state (n_(f)) . When electron in hydrogen atom jumps from energy state n_(i) = 4 to n_(f) = 3,2,1 identify the spectral series to which the emission lines belong .

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Knowledge Check

  • According to Bohr's frequency rule, the frequency of radiotion obsorbed or emitted when transtion occurs between two stationary states that differ in enegry by DeltaE is given by

    A
    `DeltaE. h^(2)`
    B
    `DeltaE//h`
    C
    `DeltaE. h^(2)`
    D
    `DeltaE.sqrt(h)`
  • If the frequency of E.M radiations is halved then the energy of EM radiation wil become

    A
    double
    B
    remains unchanged
    C
    becomes half
    D
    becomes one fourth
  • Calculate the wavelength of radiation emitted when He^(+) makes a transition from the state n =3 to the state n=2.

    A
    164.0nm
    B
    164.0 `mu` m
    C
    114.0 nm
    D
    416.0nm
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