## Q 42 on post assessment

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

Samantha Hoegl Roy 2C
Posts: 81
Joined: Fri Sep 28, 2018 12:15 am

### Q 42 on post assessment

An excited hydrogen atom emits light with a frequency of 1.14 x 1014 Hz to reach the energy level n = 4. In what principle quantum level did the electron begin?

how would you solve this question?

Jessica Castro 2H
Posts: 60
Joined: Fri Sep 28, 2018 12:29 am

### Re: Q 42 on post assessment

I believe you would use the equation for the energy level of an H- atom, but use it for the initial and final n. In this case, you would be solving for n(initial). Set this difference equal to the equation for the energy of a photon, since it takes energy to change from one energy level to another.

Samantha Hoegl Roy 2C
Posts: 81
Joined: Fri Sep 28, 2018 12:15 am

### Re: Q 42 on post assessment

can you explain it in more detail i still don't understand because i can figure out E(n=4), E of everything, and the wavelengths of E(n=4) and E(everything - n=4)

Quinn_Simpson_3D
Posts: 12
Joined: Fri Sep 28, 2018 12:26 am

### Re: Q 42 on post assessment

The Rydberg equation is 1/lambda = R((1/n^2(initial))-(1/n^2(final)), where R is Rydberg's constant. You have the frequency, which means you can find the wavelength using c=(wavelength)(frequency). Once you have the wavelength, you know the wavelength and the final electron state, which is 4. Since you have both of these numbers, you can solve for n-initial, which after you do the work is 6.

Chem_Mod
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### Re: Q 42 on post assessment

The change in energy of the electron is equal to energy of the light emitted. Having the frequency of the light allows you to calculate its energy (E=hv). You also know the energy of n=4 because E(n)=-hR/n^2. Also note that ∆E=-E(emitted light)=Ef-Ei = E(n=?) - E(n=4). With this information, you can find the other n value. Dr. Lavelle wants you to avoid using the Rydberg equation.

Pritish Patil 1K
Posts: 60
Joined: Fri Sep 28, 2018 12:24 am

### Re: Q 42 on post assessment

Use the equation for the energy level of an H-atom. (1/lambda=R((1/n^2 (initial)-(1/n^2(final))). Since we have the frequency, the lambda can be calculated. As well plug in n=4 for the final energy level and solve for the initial energy level.

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