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Physics-

General

Easy

Question

The following diagram indicates the energy levels of a certain atom when the system moves from 4E level to E. A photon of wavelength l₁ is emitted. The wavelength of photon produced during it's transition from $fraction numerator 7 over denominator 3 end fraction E$ level to E is l₂. The ratio $fraction numerator lambda subscript 1 end subscript over denominator lambda subscript 2 end subscript end fraction$ will be

$\frac{9}{4}$
$\frac{4}{9}$
$\frac{3}{2}$
$\frac{7}{3}$

The correct answer is: $fraction numerator 4 over denominator 9 end fraction$

Transition from 4E to E
$left parenthesis 4 E minus E right parenthesis equals fraction numerator h c over denominator lambda subscript 1 end subscript end fraction$ Þ $lambda subscript 1 end subscript equals fraction numerator h c over denominator 3 E end fraction$ ...... (i)
Transition from $fraction numerator 7 over denominator 3 end fraction E$ to E
$open parentheses fraction numerator 7 over denominator 3 end fraction E minus E close parentheses equals fraction numerator h c over denominator lambda subscript 2 end subscript end fraction$ Þ $lambda subscript 2 end subscript equals fraction numerator 3 h c over denominator 4 E end fraction$ ...... (i)
From equation (i) and (ii) $fraction numerator lambda subscript 1 end subscript over denominator lambda subscript 2 end subscript end fraction equals fraction numerator 4 over denominator 9 end fraction$ n the material

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