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3 years ago

What is the longest wavelength of radiation that possesses the necessary energy to break the o-o bond?

Physics

1 answer:

Vika [28.1K]3 years ago

7 0

Answer:

The longest wavelength of radiation is 241nm and it lies in ultraviolet region.

Explanation:

The minimum energy required to break o-o bond is 495kJ/mole.

The photon does not have mass and the energy of the single photon depends entirely on the wavelength and is given by

e =hc/λ

where, h is the Planck constant,

c is the speed of light

e is the energy of photon

λ is the wavelength

From the Planck formula we can understand that energy of the photon is quantized.

E = e.Nₐ

e = E/Nₐ = $\frac{495KJ}{6.022*10^{23} }$ = 8.22*10⁻¹⁹ J/photon

λ = hc/E= $\frac{6.626*10^{-34}*2.998*10^{8} }{8.22*10^{-19} }$

λ = 2.41*10⁻⁷ m = 241 nm

The longest wavelength of radiation is 241nm and it lies in ultraviolet region.

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$P(V-Nb)=NkT\\\Rightarrow P=\dfrac{NkT}{V-Nb}\\\Rightarrow P=\dfrac{3\times 6.023\times 10^{23}\times 1.38\times 10^{-23}\times 300}{0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow P=8563732.58906\ Pa$

The pressure is 8563732.58906 Pa

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$P_1(V_1-Nb)=P_2(V_2-Nb)\\\Rightarrow P_2=\dfrac{P_1(V_1-Nb)}{V_2-Nb}\\\Rightarrow P_2=\dfrac{8563732.58906(0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29})}{0.002-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow P_2=3992793.23326\ Pa$

The pressure is 3992793.23326 Pa

Work done is given by

$dw=Pdv\\\Rightarrow W=\int_{v_1}^{v_2}\dfrac{NkT}{V-Nb}dv\\\Rightarrow W=NkTln\dfrac{V_2-Nb}{V_1-Nb}\\\Rightarrow W=3\times 6.023\times 10^{23}\times 1.38\times 10^{-23}\times 300ln\dfrac{0.002-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}{0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow W=5708.00923\ J$