The Early Atom | Boundless Physics - Lumen Learning [PDF]

We have a detailed (and accurate) model of the atom now, but it took a long time to come up with the correct answer. ...

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Idea Transcript


e

m

14N + Õ 17O + p

1 fm

1 Å = 100,000 fm

10 − 10

6.022 ⋅ 1023

-

-

-

+ + ++ +

-

-

-

3.4 ⋅ 10 − 14 1 3000

10 − 10

ΔE = E2 − E 1 = h h

ΔE = E2 − E 1 = h h

L L=n

h

=n

2

n = 1, 2, 3, …

=

h 2

h

Z=1 Z>1

h n2 3Õ2

n

Z=1

L

L = mevr n = n

L

h =n 2

L

me

rn

n

r I = mr 2

L=I

h

v

=

r

v L = (mr2 )( ) = mvr r mvr

E4

n

n

rn =

n2 2

r

Zk ee2m e

− 13.6Z 2

E =»

E2

n2

eV

( ) 1

1

E = Ei − Ef = R 2 − 2 nf ni

Z

m ev 2 r

v=



Z=1

Z=2

m 2ev

Zk ee 2

rn

r2

Zk ee 2

=

r2

Zk ee 2 m er

Zk ee2 Zkee 2 1 E = mev 2 − = − 2 r 2r r r L = m erv m evr = n r



n

Zkee 2 m er = n

rn =

n2 2 Zk ee2 m e r

E= −

Zk ee 2 2r n

= −

E = Ei − E f = R

Z 2 (k ee2 ) 2 m e 2 2

2 n

n

»

−13.6Z 2 n

2

eV

( ) 1

1 − n2f n 2i

R

1

1

= R(

1 n 2f



1 n 2i

)

1 1 = R( 2 − 2 ) nf ni R

R = 1.097 ⋅ 10 7 \text{m} − 1

nf ni nf = 1

nf = 2

nf = 3 nf nf = 2

n i = 3, 4, 5, 6…

12

2 n m 10 3 n

m

nf

9

n 7

9

m

5

m 6 5 6 n 4 8 6 n m

187

ni

94

nm

n = 1 n = 2

ni

n m

434 nm

Lyman series

41 0 nm Balmer series

5 n

m 12 8

n = 3 n = 4 n = 5

2 n

m 10

94

n m

Paschen series

n = 6

nf

n = 2r h p

= h/m ev nh = 2rn m ev L = mvr

h L = mevr n = n , (n = 1, 2, 3…) 2

¢

− =

h (1 − cos) m ec ¢

h

c 2.43 ⋅ 10 − 12 = 180

`



hc

= E = 0.0228

‘−

=0

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