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BARSIC [14]
3 years ago
10

Calculate the shortest wavelength of the electromagnetic radiation emitted by the hydrogen atom in undergoing a transition from

the n = 6 level.
Chemistry
1 answer:
pav-90 [236]3 years ago
8 0

Answer:

\lambda = 9.376*10^{-8} m

Explanation:

From the question we are told that:

 n=6level

Generally the equation for Energy is mathematically given by

 E=\frac{hc}{\lambda}

Since

Energy difference will be maximum when electron return to ground state

And Shortest wavelength is emitted when  there is largest energy difference

Therefore

 1/\lambda = R* (\frac{1}{nf^2} - \frac{1}{ni^2})

Where

 R=Rydberg\ constant

 R = 1.097*10^7

Therefore

 1/\lambda = 1.097*10^7* ((\frac{1}{1^2} - \frac{1}{6^2})

  \lambda = 9.376*10^{-8} m

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Answer:

7.21 × 10⁴ J

Explanation:

Ethanol is solid below -114.5°c, liquid between -114.5°C and 78.4°C, and gaseous above 78.4°C.

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We need to calculate the heat required in different stages and then add them.

The moles of ethanol are:

48.3g.\frac{1mol}{46.07g} =1.05mol

Solid-liquid transition

Q₁ = ΔHfus . n = (4.60 kJ/mol) . 1.05 mol = 4.83 kJ = 4.83 × 10³ J

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ΔHfus: molar heat of fusion

n: moles

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Q₂ = c(l) . m . ΔT = (2.45 J/g.°C) . 48.3g . [78.4°C-(-114.5°C)] = 2.28 × 10⁴ J

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ΔT: change in the temperature

Liquid-gas transition

Q₃ = ΔHvap . n = (38.56 kJ/mol) . 1.05 mol = 40.5 kJ = 40.5 × 10³ J

where,

ΔHvap: molar heat of vaporization

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Q₄ = c(g) . m . ΔT = (1.43 J/g.°C) . 48.3g . (135.3°C-78.4°C) = 3.93 × 10³ J

where

c(g): specific heat capacity of the gas

Total heat required

Q₁ + Q₂ + Q₃ + Q₄ = 4.83 × 10³ J + 2.28 × 10⁴ J + 40.5 × 10³ J + 3.93 × 10³ J = 7.21 × 10⁴ J

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