Question

23. What is the frequency of light emitted when an
electron in a hydrogen atom jumps from the 3rd
orbit to the 2nd orbit?​

Answer 1

Explanation:

The energy of the photon emitted shall equal to the difference between the electron's potential energy at the two orbits.

The

Rydberg Formula

−−−−−−−−−−−−−−− suggests that an electron that travels (or literally "falls") from energy level

n

i

to

n

f

(

n

i

>

n

f

) emits a photon of wavelength

λ

for which

1

λ

=

R

(

1

n

2

f

−

1

n

2

i

)

where the Rydberg's Constant

R

=

1.097

×

10

7

l

m

−

1

.

In this particular scenario

n

i

=

3

n

f

=

1

Such that

1

λ

=

R

(

1

n

2

f

−

1

n

2

i

)

Δ

PE

=

1.097

×

10

7

l

m

−

1

⋅

(

1

1

2

−

1

3

2

)

Δ

PE

=

9.751

⋅

10

7

l

m

−

1

λ

=

1.026

×

10

−

7

l

m

⋅

10

9

l

n

m

1

l

m

λ

=

102.6

l

n

m

References

"Bohr's Hydrogen Atom" , Chemistry Libretext

Answer 2

Answer:

Explanation:

Lambda=-1.51-(3.4)= 1.89. n= c/lambda.