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Naya [18.7K]
2 years ago
5

Different isotopes of the same element emit light at slightly different wavelengths. A wavelength in the emission spectrum of a

hydrogen atom is 656.45 nm; for deute- rium, the corresponding wavelength is 656.27 nm. (a) What minimum number of slits is required to resolve these two wavelengths in second order
Chemistry
1 answer:
Inessa [10]2 years ago
7 0

Different isotopes of the same element emit light at slightly different wavelengths, the minimum number of slits  is mathematically given as

N=1820slits

<h3>What minimum number of slits is required to resolve these two wavelengths in second-order?</h3>

Generally, the equation for the wave is mathematically given as

d\ sin\ (\theta\ m) \ = \ m\  \lambda

Where the chromatic resolving power (R) is defined by

R\  =\ \lambda\ / \ d \  \lambda

R = nN,

Therefore

\lambda_1 \ = \  (656.45)(1 \ * \ 10^{-9})/1mm

\lambda_1= 656.45*10^{-9}

and

\lambda_2= (656.27)(1*10^{-9})/1mm

\\\\\lambda_2= 656.27*10^{-9}m

In conclusion, the minimum number of slits is required to resolve these two wavelengths in second-order

N\ =\ \dfrac{\lambda}{m\ d\ T\ }\\\\

Therefore

N\ =\ \dfrac{656.45 \ * \ 10^{-9}}{2\ * \ (0.18*10^{-9})}

N=1820slits

Read more about slits

brainly.com/question/24305019

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Answer:

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NO(g) + H₂O(g) → NH₃(g) + O₂(g)

To balance the equation, we add the stoichiometric coefficients (4 for NH₃ and NO to balance N atoms, then 6 for H₂O to balance H atoms and then 5 for O₂ to balance O atoms):

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All reactants and products are in the gaseous phase, so the equilibrium constant is expressed in terms of partial pressures (P) and is denoted as Kp. The Kp is expressed as the product of the reaction products (NH₃ and O₃) raised by their stoichiometric coefficients (4 and 5, respectively) divided into the product of the reaction reagents (NO and H₂O) raised by their stoichiometric coefficients (4 and 6, respectively). So, the pressure equilibrium constant expression is written as follows:

Kp = \frac{P(NH_{3}) ^{4} P(O_{2}) ^{5}}{P(NO) ^{4} P(H_{2}O)^{6}}

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