1.

The figure shows an energy level diagram for the hydrogen atom. Several trasitions are markedas I,II,III, …… The diagram is onlyindicative and not to scale. Then, (##MTG_WB_JEE_CHE_MTP_01_E01_040_Q01.png" width="80%">

Answer»

The transition in which a Balmerseries photon absorbed is VI.
the wavelengthof TE radiation involved in transitionII is 486 nm.
transition IV will occur when a HYDROGEN atom is irradiatedwith radiation of wavelength103 nm.
transition IV will emitthe longestwavelength line in the visibleportion of the hydrogen spectrum.

Solution :For Balmer series, ` n_(1) =2, n_(2)= 3,4….`
In transition (VI), photon of Balmerseries is absorbed.
In transitionII
` E_(2) = -3.4 eV, E_(4)= 0.85 eV, DeltaE =2 .55 eV`
` DeltaE =(12400 eV Å)/( 1030Å)=12.0 eV`
So difference of energyshouldbe 12.0 eV (approx)
Hence, ` n_(1) = 1 and n_(2)= 3 `
`(-13.6 eV) (-1.51 eV) THEREFORE `Transition is V.
For longest wavelength, energy difference should be minimum
So in visibleportionof hydrongen atom, minimum energy emitted is intransition IV.


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