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Mamont248 [21]
3 years ago
5

A hydrogen atom is in its fifth excited state. The atom emits a wavelength photon. Determine the maximum possible orbital angula

r momentum of the electron after emission. Express your answer as multiples of hbar
Chemistry
1 answer:
forsale [732]3 years ago
8 0

Answer:

(√ 20) hbar.

Explanation:

The first thing to do is to calculate the second principal quantum number, n from the formula for Calculating energy,E and this is done as given below;

Energy, E = 13.6 (1 / n^2 - 1/6).

(Where energy,E = 1240/ 7474).

Hence, when we solve the above equation for n, the principal quantum number = 5.

We know that the azimuthal quantum Number, l = (n- 1); 5 - 1 = 4.

Therefore, the maximum possible magnitude of the orbital angular momentum of the atom after emission = (√ 4 × 5) hbar.

the maximum possible magnitude of the orbital angular momentum of the atom after emission = (√ 20) hbar.

Note that hbar = 1.0545718 × 10-34 m2 kg / s.

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Answer: The new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

Explanation:

Given: V_{1} = 61 L,      T_{1} = 183 K,      P_{1} = 0.60 atm

At STP, the value of pressure is 1 atm and temperature is 273.15 K.

Now, formula used to calculate the new volume is as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{0.60 atm \times 61 L}{183 K} = \frac{1 atm \times V_{2}}{273.15 K}\\V_{2} = 54.63 L

Thus, we can conclude that the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

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