1.

According to law of photochemical equivalence the energy absorbed (in ergs/mole) is a given as (h = 6.62 xx 10^(-27) ergs , c = 3 xx 10^(10) cm s^(-1) , N_(A) = 6.02 xx 10^(23) mol^(-1))

Answer»

`(1.196 xx 10^(4))/(LAMBDA)`
`(2.859 xx 10^(5))/(lambda)`
`(2.859 xx 10^(16))/(lambda)`
`(1.196 xx 10^(16))/(lambda)`

Solution :Acc., to stark EINSTEIN's law of photochemical EQUIVALENCE
`E= (hc)/(lambda) xx N_(A)`
` E = (6.62 xx 10^(-27) xx3 xx 10^(10) xx 6.02 xx 10^(23))/(lambda)`
`E= (1.196 xx 10^(8))/(lambda)` .


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