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)`)

Options

A. `(1.196xx10^(8))/(lambda)`
B. `(2.859xx10^(5))/(lambda)`
C. `(2.859xx10^(16))/(lambda)`
D. `(1.196xx10^(16))/(lambda)`

Answer

A

Explanation

Acc. to stark Einstein.s law of photochemical equivalence
`E=(hc)/(lambda)xxN_(A)`
`E=(6.62xx10^(-27)xx3xx10^(10)xx6.02xx10^(23))/(lambda)`
`E=(1.196xx10^(8))/(lambda)`