Question

An energetically excited hydrogen atom has its electron in the 5f subshell. The electron drops down to the 3d subshell, releasing a photon in the process?

Answer 1

a) Give the n and l quantum numbers for both subshells, and give the range of possible ml quantum numbers?

b)What is the wavelength of the light emitted in the process?

Answer 2

@dumbsearch2 Hey, I was hoping that you could explain the concept and method to solve a problem like this ?

Answer 3

c) The atom now has a single electron on the 3d subshell. What is the energy required in Kj/mol required to remove this electron?

@dumbcow I did a) and b) can you help me with c) plz?

Answer 4

use the rydberg equation for energy calculations

Answer 5

Alright.

let n1 = 3
let n2= 5
R = 1.097 *10^7/m

so \[\frac{ 1 }{ \lambda} = R_\infty [ \frac{ 1 }{ n1^{2}} - \frac{1}{n2^{ 2 } }] \]
\[\frac{1}{\lambda} =R_\infty [ \frac{1}{3^{2}} - \frac{1}{5^{2}}]\]

\[\frac{1}{\lambda} = R_\infty [\frac{1}{9} - \frac{1}{25}] \]

\[\frac{1}{\lambda} = R_\infty [\frac{16}{225} ]\]
\[\lambda = \frac{1}{R_\infty [\frac{16}{225}]} \]
\[\lambda = \frac{225}{1.097 * 10^{7} * 16 } \]
\[\lambda = 1.282 * 10^{-6}\]


\[\textit{That was part b}\]

\[\textbf{So please explain a and c}\]


@aaronq

Answer 6

"Give the n and l quantum numbers for both subshells, and give the range of possible ml quantum numbers?"

it tells you in the question, "5f subshell. The electron drops down to the 3d subshell"

n is the first number
determining l: s=0, p=1, d=2, f=3

Principal quantum number, n; Azimuthal quantum number, l.
n=5 l=3
n=3 l=2


c) The atom now has a single electron on the 3d subshell. What is the energy required in Kj/mol required to remove this electron?

set up the rydberg equation: \( \dfrac{1}{\lambda}=E_{photon}=R_0(\dfrac{1}{\infty}-\dfrac{1}{3^2})\)

multiply by avogadros number to get the amount of energy per mole

Answer 7

sorry i wrote it wrong it should be: \(E=R_0(\dfrac{1}{3^2}-\dfrac{1}{\infty})\)

Answer 8

alright thanks @aaronq

like this but what is the symbol

\[\infty\]

mean?

Answer 9

it's called lemniscate, it signifies infinity. Just know that \(\dfrac{1}{\infty}=0\)

Answer 10

so basically
\[E = 1.097 * 10^{7}nm^{-1}[\frac{1}{3^{2}} - \frac{1}{\infty} ]\\
E = 12188888.89 \]

I am confused about some subjects.

In once case you can calculate the wavelength and in my textbook it says the constant is \[ R_\infty = 1.097 *10^{-2} nm^{-1}\]
in another case my answer use the constant

\[ R_\infty =1.097 *10^{7}\]

which one is right or are they both correct for different contexts.@aaronq

Answer 11

They're in different units, both are correct. the \(10^7\) is in /meters.

Answer 12

but, what do they mean. Why use one unit and not the other. Where is it derived from? Any text book reference here? By the way, have you done Rate Law pseudo constants?
@aaronq

Answer 13

Often, wavelength is expressed in nanometers, but the SI unit for distance is the meter.

1 nm = \(10^{-9}\) m

look in your textbook if you want references

Answer 14

do you mean pseudo first order, or pseudo zeroth order?

Answer 15

yes like for example
so both are first order now let
\[R = k[A][B
\\k'=k[B]=
R=k'[A]\]

Answer 16

Then we have to graph it using excel and find the best line of fit which I don't know how to do ...@aaronq any clue which function helps me convert Absorbance

\[A= \in l c \]

into concentration. Than graph it with a concentration vs time graph. that should fit \[\ln[A] = -k't + \ln[A_0]\]

or perhaps second order.

Answer 17

any ideas @aaronq

Answer 18

i'm guessing you have [B] constant?

yeah, use the beer lambert law to find concentration.

in excel you can choose line of best fit (also called "trend line").

if you're doing pseudo first order, you're basically doing first order so that would be the right equation to use.

Answer 19

Well for our Lab we actually used many different situations. We also played with the parameter of temperature where we did it once with a high temperature >30 and a low one less than < 20. There was so many to plot but If my information is no sound than I think it will probably be because of what? any ideas what could go wrong because it was really straight forward when I did the procedure.... what could I say to negate the loss of marks.? @aaronq