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3 years ago

What is the maximum number of f orbitals that are possible in a given shell?

Physics

1 answer:

Lemur [1.5K]3 years ago

7 0

Answer: seven f orbitals

Explanation:

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Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago

A balloon is filled to a volume of 7.00*10^2 mL at a temperature of 20.0°C. The balloon is then cooled at constant pressure to a

wolverine [178]

The final volume of the gas is 238.9 mL

Explanation:

We can solve this problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas (V) is proportional to its absolute temperature (T):

$\frac{V}{T}=const.$

Which can be also re-written as

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where

$V_1, V_2$ are the initial and final volumes of the gas

$T_1, T_2$ are the initial and final temperature of the gas

For the gas in the balloon in this problem, we have:

$V_1 = 7.00\cdot 10^2 mL = 700 mL$ is the initial volume

$T_1=20.0^{\circ}C+273=293 K$ is the initial absolute temperature

$V_2$ is the final volume

$T_2 = 1.00\cdot 10^2 K = 100 K$ is the final temperature

Solving for $V_2$ ,

$V_2 = \frac{V_1 T_2}{T_1}=\frac{(700)(100)}{293}=238.9 mL$

Learn more about ideal gases:

brainly.com/question/9321544

brainly.com/question/7316997

brainly.com/question/3658563

#LearnwithBrainly

6 0

3 years ago

Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa. It is observed that the amount of liquid in the cook

jeka57 [31]

Answer: a) Mr = 2.4×10^-4kg/s

V = 34.42m/a

b) E = 173J

Ø = 2693.1J

c) Er = 0.64J/s

Explanation: Please find the attached file for the solution

3 0

3 years ago

If a string vibrates at the fundamental frequency of 528 Hz and also produces an overtime with a frequency of 1,056 Hz this over

Romashka-Z-Leto [24]

I have a hunch that you're not talking about overtime
OR overtune. I think you're going for overtone .

For a fundamental frequency of 528 Hz,
1,056 Hz is the second harmonic.

5 0

3 years ago

Which of the following is the main evidence of life in the early universe?

Arturiano [62]

A) Dinosaur fossils because they were proven that there was life before us. we also found out that our old skulls had dinosaur teeth in the heads. that showed us that we were on the food chain. but now we aren't because we now have protected cities.

8 0

3 years ago

Read 2 more answers