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Structure of Atom question

2013 · Q104
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Structure of Atom question

2013 · Q104

NEETChemistryStructure of AtomMCQ+4 / −1
According to law of photochemical equivalence the energy absorbed (in ergs/mole) is given as (h = 6.62 ×\times× 10−-−27 ergs, c = 3 ×\times× 1010 cm s−-−1, NA = 6.02 ×\times× 10−-−23 mol−-−1)
  1. A
    1.196×108λ{{1.196 \times {{10}^8}} \over \lambda }λ1.196×108​
  2. B
    2.859×105λ{{2.859 \times {{10}^5}} \over \lambda }λ2.859×105​
  3. C
    2.859×1016λ{{2.859 \times {{10}^{16}}} \over \lambda }λ2.859×1016​
  4. D
    1.196×1016λ{{1.196 \times {{10}^{16}}} \over \lambda }λ1.196×1016​
View written solutionFree

Correct answer: A

E = hcNAλ{{hc{N_A}} \over \lambda }λhcNA​​

= 6.62×10−27×3×1010×6.02×1023λ{{6.62 \times {{10}^{ - 27}} \times 3 \times {{10}^{10}} \times 6.02 \times {{10}^{23}}} \over \lambda }λ6.62×10−27×3×1010×6.02×1023​

= 1.1955×108λ{{1.1955 \times {{10}^8}} \over \lambda }λ1.1955×108​ = 1.196×108λ{{1.196 \times {{10}^8}} \over \lambda }λ1.196×108​ ergs mol-1

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