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For a photochemical reaction ArarrB ,1.0...

For a photochemical reaction `ArarrB ,1.0xx10^(-5)` mole of B were formed on absorbing `1.2xx10^(19)` quanta each of `lamda=360 nm`. The quantum efficiency is given by

A

`0.50`

B

1

C

`0.1`

D

`0.2`

Text Solution

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The correct Answer is:
A
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For a photochemical reaction A to B, 1.0xx10^(-5) mole of B were formed on absorbing 1.2xx10^(19) photons, quantum efficiency is : (N_(A)=6xx10^(23))

Can a body have a charge of 1.0xx10^(-19) C ?

Knowledge Check

  • The Avogadro's number is 6xx10^(23) per gm mole and electronic charge is 1.6xx10^(-19)C . The Faraday's number is

    A
    `6xx10^(23)xx1.6xx10^(-19)`
    B
    `(6xx10^(23))/(1.6xx10^(-19))`
    C
    `(2)/(6xx10^(23)xx1.6xx10^(-19))`
    D
    `(1.6xx10^(-19))/(6xx10^(23))`

  • The rate constant for the reaction A rarr B is 2 xx 10^(-4) mol^(-1) . The concentration of A at which rate of the reaction is (1//2) xx 10^(-5) M sec^(-1) is -

    A
    `0.25 M`
    B
    `(1//20) sqrt(5//3) M`
    C
    `0.5 M`
    D
    None of these

  • Quantium efficiency or quantum yield (phi) of photochemical reaction is given by: phi = ("moles of the substance reacted")/("moles of photons absorbed") Absorption of UV radiation decomposes A according to the reaction 2A overset (hv) rarr B+3C The quantum yield of the reaction at 330 nm is 0.4 A sample of 'A' absorbs monochromatic radiation at 330 nm at the rate of 7.2xx10^(-3)Js^(-1) (Given N_(A)=6xx10^(23),h=6.6xx10^(-34) in S.I unit) The rate of formation of C ( mol //s) is

    A
    `1.2xx10^(-8)`
    B
    `8xx10^(-8)`
    C
    `8xx10^(-9)`
    D
    None of these

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