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The transition from the state n=4 to n=3...

The transition from the state n=4 to n=3 in a hydrogen - like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition.

A

`2 to 1`

B

`3 to 2`

C

`4 to 2`

D

`5 to 4`

Text Solution

Verified by Experts

In the Ltman eries we get energy in the UV region. In the Balmer series we get energy in the visible region. In the Paschen/Brackett/P Fund series we get energy in the IR region.
We know that `(1)/(lamda ) = R ((1)/( n _(1) ^(2)) - (1)/( n _(2) ^(2)))`
for hydrogen atom for `e ^(-)` transition from `n _(2) to n _(1).`
For 4 to 3
`E prop (1)/(lamda ) = R ((1)/(9) - (1)/(16)) = (7)/( 9 xx 16) R`
For 2 to 1, 3 to 2, and 4 to 2 we get more vaues of `1//lamda ,` that is , more energy than `4 to 3.`
IR radiation has less energy than UV radiation.
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Knowledge Check

  • In H atom ultraviolet radiation will be obtained in the transition.

    A
    `5 rarr 2`
    B
    `3 rarr 2`
    C
    `4 rarr 1`
    D
    `5 rarr 3`

  • An electron makes a transition from orbit n=4 to an orbit n=2 of a hydrogen atom.What is the wavelength of the emitted radiation?

    A
    16//4R
    B
    16//3R
    C
    3R//16
    D
    4R//3

  • Assertion : In Li^(2+) ion, an electron make transition from higher state to n=2. Then the photon observed will fall in the visible range. Reason : Line falling in n=2 is balmer series line which belongs to visible range in all type of H-like atom orion.

    A
    If both (A) and (R) are correct and (R) is the correct explanation of(A).
    B
    If both (A) and (R) are correçt, but (R) is not the correct explanation of (A).
    C
    If (A) is correct, but (R) is incorrect.
    D
    If both (A) and (R) are incorrect.

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