## Atomic Spectra Module #41

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

mahika_nayak_3L
Posts: 60
Joined: Fri Sep 28, 2018 12:26 am

### Atomic Spectra Module #41

The question that I had is from the Atomic Spectra Post Assessment Module.

#41 states For the hydrogen atom which statement is true?

A. The transition from n = 5 to n = 3 involves greater energy than one from n = 4 to n = 2.

B. The transition from n = 4 to n = 2 emits radiation of longer wavelength than the transition from n = 5 to n = 1.

C. All transitions from states for which n > 1 to the n = 1 state involve the absorption of energy by the atom.

D. A transition from n = 2 to some large value of n corresponds to the ionization energy of the H atom.

Can someone explain what the answer would be and why?

Brian Chhoy 4I
Posts: 66
Joined: Fri Sep 28, 2018 12:16 am

### Re: Atomic Spectra Module #41

Its B. You know that an electron going from 5 to 1 would emit more energy than that from 4 to 2, since the energy emitted is proportional to the distance the electron falls. Energy is also inversely proportional to wavelength, with higher energy having smaller wavelengths.

If you would want to see the math for it use the equation E=hR((1/n^2)-(1/n^2)) with the first n being the smaller number and E=hc/$\lambda$. Solve for wavelength and you get 456nm for the 4 to 2 transition and 95nm for the 5 to 1.

Karina Koo 2H
Posts: 49
Joined: Fri Sep 28, 2018 12:24 am

### Re: Atomic Spectra Module #41

#41 states For the hydrogen atom which statement is true?

A. The transition from n = 5 to n = 3 involves greater energy than one from n = 4 to n = 2.
This is false because, as elections reach higher and higher energy states, the energy gap between each state gets smaller and smaller. Therefore, even though 5-3 is moving down 2 energy states just like 4-2, the energy difference between 5-3 is smaller since they are energy states that are higher up than 4-2. That means there is less energy involved in moving from 5-3.

B. The transition from n = 4 to n = 2 emits radiation of longer wavelength than the transition from n = 5 to n = 1.
Moving from 5-1 would involve a greater amount of energy because there is a greater energy difference from 5-1 compared to 4-2. In other words, moving from 4-2 requires less energy. Less energy means longer wavelengths because wavelength is inversely proportional to energy. So, this is true!

C. All transitions from states for which n > 1 to the n = 1 state involve the absorption of energy by the atom.
If an electron is moving from a energy state that is higher than 1 to energy state 1( n=1) it is emitting energy so this is false.

D. A transition from n = 2 to some large value of n corresponds to the ionization energy of the H atom.
um... well we didn't cover anything about ionization energy in this topic to this is just unrelated. False!

Meghanhe1l
Posts: 60
Joined: Fri Sep 28, 2018 12:17 am

### Re: Atomic Spectra Module #41

The answer is B because the energy difference from n=5 to n=1 is greater than the difference from n=4 to n=2. Because E=hv and h is a constant, then the greater energy difference can be reflected only in a difference in frequency, meaning that the released energy when passing from n=5 to n=1 would have a higher frequency. Becuase frequency = c/wavelength, and c is a constant, then the higher frequency would result in a shorter wavelength.

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