B) It turns out that in a deuterium atom (hydrogen with one extra neutron) the energy levels are slightly shifted. The n-3 state is about 4.0 meV lower energy than in Hydrogen, and the n=2 state is about 8.5 meV lower energy than in Hydrogen. (meV = milli electron Volts) Find the wavelength of the light emitted by the n=3 to 2 transition in Deuterium.
B) It turns out that in a deuterium atom (hydrogen with one extra neutron) the energy levels are slightly shifted. The n-3 state is about 4.0 meV lower energy than in Hydrogen, and the n=2 state is about 8.5 meV lower energy than in Hydrogen. (meV = milli electron Volts) Find the wavelength of the light emitted by the n=3 to 2 transition in Deuterium.
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Need help with Part B. Use part A to find Part B.
Transcribed Image Text:ulation, to four significant figures! This is
needed for part B. No credit for simply using the Rydberg shortcut.)
B) It turns out that in a deuterium atom (hydrogen with one extra neutron) the energy levels are
slightly shifted. The n-3 state is about 4.0 meV lower energy than in Hydrogen, and the n=2
state is about 8.5 meV lower energy than in Hydrogen. (meV = milli electron Volts) Find the
wavelength of the light emitted by the n=3 to 2 transition in Deuterium.
C) Explain in about three sentences of detail, along with a quick sketch, an experiment you could
perform to see the slight difference between the light coming from Hydrogen (part A) and the
light coming from Deuterium (part B). No numbers or calculations are needed in this part.
Problem 4. When you plug in a device like a computer, often there is a transformer box on the
cable to lower the rms voltage to about 12 volts.
A) Explain how such a transformer would work, at the detailed level of charges/currents/fields.
(Probably will require ~3 sentenpage Give the relevant Qrameters to make sure it would end
up at 12 volts if using an ordinary outlet.
O
?
Transcribed Image Text:For
a
hydrogen atom,
Energy state is given by:
En =
13.6eu
2
n
n = 1, 2, 3 4
n=3 state.
E3 =
1-2 state E₁=
Absorbed
Energy
When
the
difference between
Energy
of
energy
he
E3-E2
K
ис
Wave length: A = E₁-E₂
A
For
For
the
-13.6ev
3²
13.6ev
2²
= -1.51 eV
=
- 3.4eV
n=u
-n= 3 E₂ = -1. Slev
- 3.4eV
n = 2
emitted
energy
election drops from n=3 to n=2, the
the states is equal to
the emitted radiation or
light.
E3-E₂= (-1.5lev)-(-3.4 eV) = 1.89 ev
Joules = 1.89 x 1.6x10 J = 3.024 × 150 195
-19
λ=
(6.63 x 10³5.5) x 3 x 10³ m/s)
(3.024x10-145)
657.3x10m=657.3
=
n
4= 6
hes
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