Chemistry- Atomic Sturcture

[xiphoid]

Homework Statement

The hydrogen electron de-excites from its 3rd excited state, which are the true and false statements for it.

1. It emits radiation giving lines in Lymann, Balmer, Paschen series
2. It emits radiation of only UV and Visible region and not Infrared region
3. It emits radiation of smallest wavelength giving Line in Lymann series
4. It will emit radiation of highest frequency giving line in pfund series.

Homework Equations

ΔE = R_h[1/R_i² - 1/R_f^s]

The Attempt at a Solution

Clearly, third excited state should mean that n=4.
So it will emit radiation in Lymann, Balmer, Paschen series, hence the statement should be true.
Then, as per the above mentioned sentence, it will emit radiation in Infrared region also, hence the statement should be false.
Smallest wavelength- will be when the difference in wavelength will be 1, eg. n_i=3, and n_f=2, if this is true, then the third statement will be true.
Highest frequency- when the difference is maximum, hence the last statement is false.

So, the answer should be TFTF.?

 

[AGNuke]

3. Smallest wavelength=Highest Energy. For Lymann series, thus corresponds to electron jump from $\infty \to 1$, which is 91.2 nm. so the third statement is false.

4. This statement is false because for Pfund series, n_f = 5.

Answer is TFFF.

 

[xiphoid]

yeah, got it! i was actually initially confused regarding that while typing, but now got it.

But while referring to the solution of this sum, they had considered the third excited state when n= 3, which i don't think is true, do you agree? it should be n=4

3. Smallest wavelength=Highest Energy. For Lymann series, thus corresponds to electron jump from $\infty \to 1$, which is 91.2 nm. so the third statement is false.

4. This statement is false because for Pfund series, n_f = 5.

Answer is TFFF.

 

[AGNuke]

3rd excited state is n=4. The solution has some typo, as mostly, solutions are not typed by teachers, but professional typists, who go out of their way to "correct" them.