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The figure shows an energy level diagram...

The figure shows an energy level diagram for the hydrogen atom. Several trasitions are marked as I,II,III, …… The diagram is only indicative and not to scale. Then,
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A

The transition in which a Balmer series photon absorbed is VI.

B

the wavelength of te radiation involved in transition II is 486 nm.

C

transition IV will occur when a hydrogen atom is irradiated with radiation of wavelength 103 nm.

D

transition IV will emit the longest wavelength line in the visible portion of the hydrogen spectrum.

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
For Balmer series, ` n_(1) =2, n_(2)= 3,4….`
In transition (VI), photon of Balmer series is absorbed.
In transition II
` 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 energy should be 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 visible portion of hydrongen atom, minimum energy emitted is in transition IV.
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