Applying bohr's model what electron of h atom comes from n=4 to n=2 calculate its wavelength
Answers
energy transition will be equal to 1.55 * 10^(-19)"J"
1.55
⋅
10
−
19
J
.
So, you know your energy levels to be n = 5 and n = 3. Rydberg's equation will allow you calculate the wavelength of the photon emitted by the electron during this transition
1/(lamda) = R * (1/n_("final")^(2) - 1/n_("initial")^(2))
1
λ
=
R
⋅
(
1
n
2
final
−
1
n
2
initial
)
, where
lamda
λ
- the wavelength of the emitted photon;
R
R
- Rydberg's constant - 1.0974 * 10^(7)"m"^(-1)
1.0974
⋅
10
7
m
−
1
;
n_("final")
n
final
- the final energy level - in your case equal to 3;
n_("initial")
n
initial
- the initial energy level - in your case equal to 5.
So, you've got all you need to solve for lamda
λ
, so
1/(lamda) = 1.0974 * 10^(7)"m"^(-1) * (1/3^2 - 1/5^2)
1
λ
=
1.0974
⋅
10
7
m
−
1
⋅
(
1
3
2
−
1
5
2
)
1/(lamda) = 0.07804 * 10^(7)"m"^(-1) => lamda = 1.28 * 10^(-6)"m"
1
λ
=
0.07804
⋅
10
7
m
−
1
⇒
λ
=
1.28
⋅
10
−
6
m
Since E = (hc)/(lamda)
E
=
h
c
λ
, to calculate for the energy of this transition you'll have to multiply Rydberg's equation by h * c
h
⋅
c
, where
h
h
- Planck's constant - 6.626 * 10^(-34)"J" * "s"
6.626
⋅
10
−
34
J
⋅
s
c
c
- the speed of light - "299,792,458 m/s"
299,792,458 m/s
So, the transition energy for your particular transition (which is part of the Paschen Series) is
E = (6.626 * 10^(-34)"J" * cancel("s") * "299,792,458" cancel("m/s"))/(1.28 * 10^(-6)cancel("m"))
E
=
6.626
⋅
10
−
34
J
⋅
s
⋅
299,792,458
m/s
1.28
⋅
10
−
6
m