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Structure of Atom question

2011 · Q87
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Structure of Atom question

2011 · Q87

NEETChemistryStructure of AtomMCQ+4 / −1
The energies E1 and E2 of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths i.e., λ\lambdaλ1 and λ\lambdaλ2 will be
  1. A
    λ\lambdaλ1 = λ\lambdaλ2
  2. B
    λ\lambdaλ1 = 2λ\lambdaλ2
  3. C
    λ\lambdaλ1 = 4λ\lambdaλ2
  4. D
    λ\lambdaλ1 = 12{1 \over 2}21​ λ\lambdaλ2
View written solutionFree

Correct answer: B

E1 = hcλ1{{hc} \over {{\lambda _1}}}λ1​hc​ and E2 = hcλ2{{hc} \over {{\lambda _2}}}λ2​hc​

E1E2{{{E_1}} \over {{E_2}}}E2​E1​​ = hcλ1×λ2hc{{hc} \over {{\lambda _1}}} \times {{{\lambda _2}} \over {hc}}λ1​hc​×hcλ2​​ = λ2λ1{{{\lambda _2}} \over {{\lambda _1}}}λ1​λ2​​

⇒\Rightarrow⇒ 255025 \over 505025​ = λ2λ1{{{\lambda _2}} \over {{\lambda _1}}}λ1​λ2​​

⇒\Rightarrow⇒ 121 \over 221​ = λ2λ1{{{\lambda _2}} \over {{\lambda _1}}}λ1​λ2​​ ⇒\Rightarrow⇒ λ1\lambda _1λ1​ = 2λ2\lambda _2λ2​

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