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Furkat [3]
3 years ago
6

What is emitted when an electron returns to its lower energy level?

Chemistry
2 answers:
Yakvenalex [24]3 years ago
4 0

Answer:

When the electron changes levels, it decreases energy and the atom emits photons. The photon is emitted with the electron moving from a higher energy level to a lower energy level. The energy of the photon is the exact energy that is lost by the electron moving to its lower energy level.

Explanation:

jok3333 [9.3K]3 years ago
4 0

Answer:

A photon

Explanation:

Hope it helps!

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0.0500 mol of gas occupies a cylinder which is sealed on top by a moveable piston. The piston is circular, with a mass of 30.0 k
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Answer:

The workdone by both N₂ and neon gas  is 49.3 J

The change in internal energy of N₂ and neon gas is 125.6 J and 73.54 J respectively

The heat  for N₂ and neon gas is 171.9 J and 122.84 J respectively.

Explanation:

Given that:

number of moles = 0.05 mole

mass of the piston = 30 kg

diameter = 5.00 cm = 0.05 m

Area (A) = πr²

Area (A) = \pi*(\frac{0.05}{2})^2 \\ \\ Area (A) = 0.0019635 \\ \\ Area (A) = 19.635*10^{-4} \ m^2

The piston is said to move from 30 cm - 40 cm

So, the change in volume ΔV is calculated as:

=(40-30)*10^{-2} *19.635*10^{-4}

= 1.9635*10^{-4} \ m^3

Outside the cylinder; the pressure P_{air}= 1 \ atm = 101325 Pa

Thus, workdone w_1 = PΔV

= 101325*1.9635*10^{-4}

= 19.90 J

The gravitational work w_2 = mgh

Given that the height (h) = 10 cm  = 0.1 m

Then;  w_2 = 30*9.8*0.1

w_2 = 29.4 \  J

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w_{total } =w_1 + w_2

w = (19.90 + 29.4) \ J

w =49.3 \ J

The pressure of gas inside the cylinder is determined as:

P_{in}.A = P_{out}.A +mg

(P_{in}-P_{out}) = \frac{mg}{A} \\ \\ P_{in} -10^5 = \frac{30*9.8}{19.635*10^{-4}} \\ \\ P_{in} = 149732.6203+10^5 \\ \\ P_{in} = 2.497*10^5 \ Pa

a). assuming that the gas is N₂.

C_v =\frac{5}{2}R

Thus, the change in internal energy ΔU is given as:

\delta U = nC_v \delta T \\ \\ \delta U = n* \frac{5}{2}R \delta T \\ \\ \delta U = \frac{5}{2}nR \delta T

Since P_{in} \delta V = nR \delta T ; \ Then;

\delta \ U = \frac{5}{2} P_{in} \delta V \\ \\ \delta \ U = \frac{5}{2}*2.497*10^5 *1.9635*10^{-4} \\ \\ \delta \ U = 122.57 \ J

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b) In Neon gas:

C_v = \frac{3}{2}R

∴

change in internal energy is;

\delta U = nC_v \delta T \\ \\ \delta U = n* \frac{3}{2}R \delta T \\ \\ \delta U = \frac{3}{2}P_{in}.V

\delta U = \frac{3}{2}*2.497*10^5*1.9635*10^{-4}

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The heat Q = ΔU + W

Q = (73.54 + 49.3) J

Q = 122.84 J

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