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Inessa [10]
3 years ago
14

An atom emits a photon of wavelength 1.08 meters. What is the energy change occurring in the atom due to this emission? (Planck'

s constant is 6.626 × 10-34 joule seconds, the speed of light is 2.998 × 108 m/s)
Chemistry
2 answers:
kati45 [8]3 years ago
4 0
The answer is 1.84×10^(-25).
igor_vitrenko [27]3 years ago
4 0

Answer : The energy change occurring in the atom due to this emission is, 18.38\times 10^{-26}J

Solution :

Formula used :

\Delta E=\frac{h\times c}{\lambda}

where,

\Delta E = change in energy of photon = ?

h = Planck's constant = 6.626\times 10^{-34}Js

c = speed of light = 2.998\times 10^8m/s

\lambda = wavelength = 1.08 m

Now put all the given values in the above formula, we get the energy of the photons.

\Delta E=\frac{(6.626\times 10^{-34}Js)\times (2.998\times 10^8m/s)}{1.08m}

\Delta E=18.39\times 10^{-26}J

Therefore, the energy change occurring in the atom due to this emission is, 18.38\times 10^{-26}J

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Explanation:

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The minimum pressure should be 901.79 kPa

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<u>Step 1: </u>Data given

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The formula for the osmotic pressure =

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6 0
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