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41 votes
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Calculate the energy of a photon of light emitted from a hydrogen atom when an electron falls from level n

User Lex Li
by
2.5k points

1 Answer

20 votes
20 votes

Answer: E

1

E

4

=

Δ

E

=

2.18

×

10

18

J

(

1

n

2

f

1

n

2

i

)

=

2.18

×

10

18

J

(

1

1

2

1

4

2

)

=

2.18

×

10

18

J

(

15

16

)

=

2.04

×

10

18

J

After you obtain the energy, then you can realize that that energy has to correspond exactly to the energy of the photon that came in:

|

Δ

E

|

=

E

photon

=

h

ν

=

h

c

λ

where

h

is Planck's constant,

c

is the speed of light, and

λ

is the wavelength of the incoming photon. Thus, the wavelength is:

λ

=

h

c

E

photon

=

(

6.626

×

10

34

J

s

)

(

2.998

×

10

8

m/s

)

2.04

×

10

18

J

=

9.720

×

10

8

m

=

97.20 nm

Step-by-step explanation:

User Manish Agrawal
by
2.7k points