## De Brogile Sapling HW Problem help

Heather Szeszulski 1I
Posts: 59
Joined: Wed Sep 30, 2020 9:55 pm

### De Brogile Sapling HW Problem help

The average speed of a diatomic nitrogen molecule at 25 ∘C is 515.2 m⋅s−1 . What is the average wavelength of a nitrogen molecule at this temperature? Assume that the molecule acts as a single particle.

I feel like there is a variable missing in order to solve this problem.. using de broglie wavelength equation can someone please help.

Vivian Leung 1C
Posts: 119
Joined: Wed Sep 30, 2020 10:09 pm

### Re: De Brogile Sapling HW Problem help

Hi Heather,

I did a similar problem but for an oxygen molecule instead.

You are using this equation right: Wavelength= planck's constant (h) divided by momentum (p)

This is the same equation as: Wavelength= planck's constant (h) divided by (mass times velocity)

Planck's constant= 6.63 x 10^-34 J.s
From the problem, you know that the speed (velocity) of the molecule: is 515.2 m/s
The molecule that you are dealing with is a diatomic nitrogen molecule. The question asks you to treat the molecule as acting as a single particle. Therefore, the mass of your molecule can be found by looking at nitrogen's atomic mass on the periodic table and calculating the total mass (make sure it is in kg) for diatomic nitrogen (n2).

The wavelength can then be found through the De Broglie equation.

Heather Szeszulski 1I
Posts: 59
Joined: Wed Sep 30, 2020 9:55 pm

### Re: De Brogile Sapling HW Problem help

Hi thats what I tried to do but I keep getting it wrong for the mass of the nitrogen particle would it be 4.652*10^-23??

Yu Jin Kwon 3L
Posts: 61
Joined: Wed Sep 30, 2020 9:41 pm
Been upvoted: 1 time

### Re: De Brogile Sapling HW Problem help

Hi all!
My Sapling question involved a diatomic fluorine molecule going 442.4 m/s.

While I was looking back on this problem to try and help you, I realized that I first approached it using E = (0.5)mv^2, with mass being 37.996 x 10^-3 kg (mass of diatomic fluorine molecule) and velocity being 442.4 m/s. From this, I got E = 3718.25 J. Then, I set this equal to hc/lambda to solve for wavelength, from which I got 5.33 x 10^-29. After checking this answer in the Sapling homework, I realized I got the wrong answer. Then, I approached this problem the "de Broglie way" and got the correct answer (2.37 x 10^-11 m/s).

Can someone explain why you can't approach this problem the E = (0.5)mv^2 way? I feel like there are so many equations where things are equal to E but I'm unsure if the E is for the energy of photons or the energy of electrons.

Yu Jin Kwon 3L
Posts: 61
Joined: Wed Sep 30, 2020 9:41 pm
Been upvoted: 1 time

### Re: De Brogile Sapling HW Problem help

Hi Heather! Sorry I saw your new question after I uploaded my response.

I didn't have your exact question (mine dealt with a diatomic fluorine molecule), but it seems as though the mass of the diatomic nitrogen molecule should be 28.01g/mol or 28.01 x 10^-3 kg/mol. Then, I believe you do get 4.652 x 10^-23 g/particle. But make sure to convert that to kg when you plug it back into de Broglie's because it has Planck's constant, which has J in it (J uses kg in its units).

Vivian Leung 1C
Posts: 119
Joined: Wed Sep 30, 2020 10:09 pm

### Re: De Brogile Sapling HW Problem help

Can someone explain why you can't approach this problem the E = (0.5)mv^2 way? I feel like there are so many equations where things are equal to E but I'm unsure if the E is for the energy of photons or the energy of electrons.

I am not a hundred percent sure but I think that the E= (0.5) mv^2 calculates the Kinetic energy (which from what Dr. Lavelle said before in terms of the photoelectric effect is equal to the Energy of the photon minus the work function).

Melanie Lin 3E
Posts: 55
Joined: Wed Sep 30, 2020 9:38 pm

### Re: De Brogile Sapling HW Problem help

Vivian Leung 2D wrote:Can someone explain why you can't approach this problem the E = (0.5)mv^2 way? I feel like there are so many equations where things are equal to E but I'm unsure if the E is for the energy of photons or the energy of electrons.

I am not a hundred percent sure but I think that the E= (0.5) mv^2 calculates the Kinetic energy (which from what Dr. Lavelle said before in terms of the photoelectric effect is equal to the Energy of the photon minus the work function).

Hi Vivian!
i think the formula you mentioned shows the Kinetic energy released from the metal surface is the energy of the electron. Looking back at my notes from the textbook m is actually the mass of an electron and it makes sense since during the experiment of the photoelectric effect, the electron is released from the metal surface after being hit by the light.

Shannon Moore 2L
Posts: 58
Joined: Wed Feb 26, 2020 12:17 am

### Re: De Brogile Sapling HW Problem help

Hello! You can find the mass needed for the DeBroglie equation by using the molar mass of nitrogen and Avogadro's constant.

28.01 x 10^3 kg/mol / 6.022 x 10^23 mol^-1

Then use this answer for your mass in the equation! Hope this helped :)