This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Name ID First Midterm  Chem 20A Fall 2008
Read all three questions before answering. They are NOT ordered by difficulty. Use only pencils and calculators; anything else (books, notes, etc.) are not allowed.
Don’t spend unnecessary time on mathematical manipulations. Useful info: h : 610'341 s v : V / X c = 300,000 km/sec
me: 9.110‘3‘kg
1eV= 1.6 ~10‘19J
1 mol 2 61023 ”hp : h p=mv density(onee1ectron—inorbital‘P—at —position _ r ) = ‘I’(r)2 E(kinetic) = % mv2 E(total, Hydrogenlikeatom) = —zz 13.6eV / n2
hzn2 E(total, 1D article in abox =
p ) 8mL2 1) Say we conduct a photoelectric—effect experiment with light of a given frequency, and ﬁnd,
that for that frequency, we can stop the electron ﬂow by applying on the sample a reserve voltage
of at least 1.25V. Assuming we do not apply the reserve voltage then, for that given frequency: a) What’s the maximum velocity of the emitted photoelectrons 1.25 V A IZSCV CEway cad. e/Eclwn {5215)
K. E : /25e,V f M) :2 Gan/”J—
/~ cV ' Kg; :1 Me \fz : 2.0mm” U— 2
v  W
9‘ I ”/0'3'19 V: {(3,000 {$2. —; C63X/0’c‘? M b) What’s the mini mum Awavelength of these electrons? I? : {A70 0:5 2:}? .— ev (7.1% 315)(4;3x/o‘g4) 2: 7.7 9 wow/n *‘ 9.97/4/ 7)
a) Let’s assume that the 18 orbital has the following mathematical form (This is not the correct
form, but it is close enough for our purposes here). ‘I’ls (x,y,z) : 0.91 A”
1+4(r/X)2
where r is the spherical coordinate which is the distance from the origin (r = sqrt(x2+y2+zz)). Calculate the probabigty to ﬁnd Electrons in He in the box 2
—1 .1 <x<0.9 _ / J
0.95 A<y<105 A Pmb ‘ >{5/ V
0.95 A<z<105 A 7’rolo {Hal’5 2 His/24X A)’ AZ
Maw”? =W : 4r? Ax: 0.2A
4y :: 0.] :40
42“— C7.”1 Prob (He) : Z ”[341 Emma/2470M“) 2x05
/+ W?) b) We are given a nano rod which is effectively a 1D particle in a box. Given that the ground to
the ﬁrst excited state energy is 1 eV, then what is the energy needed to excite an electron from the
ground to the second excited state? Ground 5am“ nc/ ~1. where, A . )
En : L1 ”‘2’ .‘: Aﬁl (,3 50W andan‘l‘ OF'I'I‘IJ' JXW
8m]. ): A(z"'/’) ’ 37" “/CV 3) When methane (CH4) burns in air, the temperature rises; this can cause a reaction whereby
oxygen molecules 1’ 02) and nitrogen molecules (N2) react to form N02. Say we burn l. 0 gram of methane 1n a container which has 6 0 grams of oxygen and because of
that the remainder of the 02 (after burning of the methane) reacts with N2. How many grams of N02 will be produced?
ENC/4'1 #202 ‘~> C02 79/420
' I”? H?
32. 0
_ C217! lmp/ Ch:) 2 mo} 02 J<3______ 9 2): ‘ r
"a? 759 0/ [ /me)CHyJ /mo/0 %09m02 [131/9911] 0,.) —— (02. buwn‘d) : (02 /e.,c+)
60 gram; 02 — L/roﬂrams 02 3 2,03%: 01 ...
View
Full Document
 Fall '05
 Scerri

Click to edit the document details