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. `(1.196 xx 10^(4))/(lambda)`
B. `(2.859 xx 10^(5))/(lambda)`
C. `(2.859 xx 10^(16))/(lambda)`
D. `(1.196 xx 10^(16))/(lambda)`

Explanation

To solve the problem according to the law of photochemical equivalence, we need to calculate the energy absorbed in ergs per mole using the given constants. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Formula
The energy (E) absorbed per mole can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \times N_A \]
where:
- \( h \) = Planck's constant
- \( c \) = speed of light
- \( \lambda \) = wavelength (in cm)
- \( N_A \) = Avogadro's number

### Step 2: Substitute the Values
We are given:
- \( h = 6.62 \times 10^{-27} \) ergs
- \( c = 3 \times 10^{10} \) cm/s
- \( N_A = 6.02 \times 10^{23} \) mol\(^{-1}\)

We need to substitute these values into the formula. However, we need the wavelength (\( \lambda \)) to proceed. Assuming a wavelength is provided or can be assumed for the calculation, let’s denote it as \( \lambda \).

### Step 3: Calculate the Energy
Substituting the values into the equation:
\[ E = \frac{(6.62 \times 10^{-27} \text{ ergs}) \times (3 \times 10^{10} \text{ cm/s})}{\lambda} \times (6.02 \times 10^{23} \text{ mol}^{-1}) \]

### Step 4: Simplify the Expression
Now, we can simplify the expression:
\[ E = \frac{(6.62 \times 3) \times 10^{-27 + 10} \times 6.02 \times 10^{23}}{\lambda} \]
\[ E = \frac{19.86 \times 10^{-17} \times 6.02 \times 10^{23}}{\lambda} \]
\[ E = \frac{119.6 \times 10^{6}}{\lambda} \text{ ergs/mol} \]

### Step 5: Final Calculation
Assuming \( \lambda \) is known, you can substitute it in to find the final energy value in ergs per mole. If we assume a specific wavelength, for example, \( \lambda = 5000 \) Å (or \( 5 \times 10^{-5} \) cm), we would calculate:
\[ E = \frac{119.6 \times 10^{6}}{5 \times 10^{-5}} \]
\[ E = 2392 \times 10^{11} \text{ ergs/mol} \]
Convert this to a more manageable number:
\[ E = 2.392 \times 10^{14} \text{ ergs/mol} \]

### Conclusion
The energy absorbed per mole according to the law of photochemical equivalence can be calculated as shown, and the final answer will depend on the specific wavelength used in the calculation.