White phosphorous is a tetra atomic solid $(P_{4})$ at room temperature given $Δ H_{s u b l i m a t i o n}$ of $P_{4} (s) = 59 K J / m o l$
$P_{4} (s) = 59 K J / m o l$ of $P_{4} (g) = 1265 K J / m o l$ Then wavelength of light capable of breaking P-P bond is $x \times 10^{- 7}$ meter then is

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Detailed Solution

$P_{4}$ has 6 P-P bonds

$\begin{array}{l} Δ H_{n e t} = Δ H_{atomisation} + Δ H_{sublimation} \\ = 1265 + 59 = 1324 {KJmol}^{- 1} \\ But Δ H_{net} = 6 E_{P - P} \\ 6 E_{p - p} = 1324 \\ E_{p - p} = \frac{1324}{6} = 201 KJ / mol \end{array}$

$E = \frac{h c}{λ} \Rightarrow λ = \frac{6.625 \times 10^{- 34} \times 3 \times 10^{8} \times 6 \times 10^{23}}{201 \times 10^{3}}$

$= 5.94 \times 10^{- 7} m$

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