Question

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Answer 1

Question:

0₂ undergoes photochemical dissociation into one normal oxygen atom and one oxygen atom 1.967 V more energetic than normal. The dissociation of O, into two normal atom of oxygen requires 498 kdmol. What is the maximum wavelength effective for photo chemical dissociation of 0₂ ?

ㅤㅤㅤㅤㅤOR

O₂ undergoes photo chemical dissociation into 1 normal oxygen atom (0) and more energetic oxygen atom (0*). If O* has 1.967 eV more energy than O and normal dissociation energy of O, is 498 kJ mol-¹. What is the maximum wavelength effective for the Photo Chemical dissociation of O₂?

Answer:

$★ \textbf{Dissociation energy = } 498 { \text{ \: kj \: mol} }^{ - 1}$

$= \frac{498 \times 100}{6.02 \times {1 0}^{23} } {\text{ \:J \: molecule } }^{ - 1}$

$= 8.27 \times {10}^{ - 19} {\text{ \:J \: molecule } }^{ - 1}$

$★\text{Excitation energy to form O*} = 1.967 \: {\text{eV \: } \text{atom} }^{ - 1}$

$= 1.967 \times 1.602 \times {10}^{ - 19} \: \text{J} \: {\text{atom} }^{ - 1}$

$= 3.15 \times {10}^{ - 19} \: \text{J} \: {\text{atom} }^{ - 1}$

$\textbf{★Total Energy = } 8.27 \times {10}^{ - 19} + 3.15 \times {10}^{ - 19} \text{J}$

$= 11.42 \times {10}^{ - 19}\text{J}$

$∴E = \frac{hc}{λ}$

$⇒λ = \frac{hc}{E} = \frac{6.625 \times {10}^{ - 34} \text{J} \times 3 \times {10}^{8} m}{11.42 \times {10}^{ - 19} \text{J}}$

$= 174 \times {10}^{ - 9} m = 174 \: nm$

$\\ \\ \\ \\ \sf \colorbox{lightgreen} {\red★ANSWER ᵇʸɴᴀᴡᴀʙ}$