# Maple TA 26_3 ## Maple TA 26_3

Does anyone know how to do question 2 on this quiz I am really confused!!

sjames
Guest

## 26-3 #1

The answer is 1 b/c the quantum state is 0.

## 26-3 #2

Use equation 26.8 (Zeeman Effect). This is a simple plug-in calculation.

e = charge of an electron (1.6E-19)
h = Planck's constant (6.6E-34)
B = magnetic field
m = mass of an electron (9.1E-31)

The answer we get is in Joules. Convert this to eV, and that is the final answer.

## 26-3 #3

l = n-1

Plug in your given principal number to find l.

Then, as sheet 46' shows, lay out the possible orbital states for all numbers of l, from 0 --> n-1. Then, count the quantum states available in the highest value of l.

So, for example: n=3

l = n-1 = 3-1 = 2

l=0 --> 0
l=1 --> -1, 0, +1
l=2 --> -2, -1, 0, 1, 2

Since l=2 is the highest orbital momentum level, there are 5 states available. That is the answer.

## 26-3 #4

Procedure is shown on sheet 55.

Find l, and then list out all possible states for that all. Then, assume each possible state has 2 electrons. That is your answer.

## question 2

I have tried several times to plug in the numbers into the equation for number 2 do you think you could post a more detailed explanation of how you did it because I keep getting the wrong answer.

sjames
Guest

## question 2

I dont understand number 2 either if someone could please explain I would really appreciate that.

hwilson
Guest

## Re: Maple TA 26_3

hwilson wrote:I dont understand number 2 either if someone could please explain I would really appreciate that.

Alright. The equation is:

E = ml * ((e*h*B)/(4*pi*m))

Now, plug in numbers:
ml = Number given.
e = charge of an electron (1.6E-19)
h = Planck's constant (6.6E-34)
B = magnetic field given
m = mass of an electron (9.1E-31)

Solve, and it gives you the energy in conventional units. Convert this into electron volts (eV). This is the answer.

## having prob with # 1

hey guys, im' having a real problem figuring out how to get # 1
heres the question once again:
The quantum mechanical electron in the hydrogen atom has the orbital (angular momentum) quantum number l = 6. How many quantum states are available to the electron (sheet 45) ? (Disregard electron spin)

phyfrk
Guest

## # 1

can someone please explain how to answer question number 1. what are the steps behind the answer?

physics!
Guest

## #1

same exact way as #3 if I=4 then ur answer would be 9 because
-4,-3,-2,-1,0,1,2,3,4

ksjhdhd
Guest

## Q#2, #4

Q#2.

I used the equation given above but I still can't get the right answer. my quantum number m = -3 which equal ml?

Q#4.

I can't get that either. Must not be my day:)

phent
Guest

## number 4

i'm pretty sure the answer is to take your n and subtract 1 to get your l. from here, you need to know the structure of an atom, or follow this chart:
0= s = 2 electrons
1= p= 6 "
2= d= 10 "
3= f= 14 "
4= g= 18 "
5= h= 22 "
6= i= 26 "

so my problem:n=4
n-1=1 therefore 4-1=3
follow 3 across on the chart and get 14, which was my answer.

positive.

hope it works for everyone.

number 4
Guest

## #2

how do u conver to eV because I am not getting it.

cutiepie
Guest

## Re: Maple TA 26_3

1 eV = 1.60217646E-19 J

JJ
Guest

## Re: Maple TA 26_3

I still can't get #2

can u guys do ur problems and plug it in

cause i keep getting different answers and all wrong

i converted by dividing the answer by 1.6E-19

dot
Guest

## Re: Maple TA 26_3

i got it nevermind guys

dot
Guest

## Re: Maple TA 26_3 