Nanophysics sheet (1)
1. What is the de Broglie wavelength of a photon with a momentum 7.6 × 10-35 kg∙m/s?
Estimate h as 6.63 × 10-34J s
6.15 eV
6.88 eV
7.25 eV
8.24 eV
8.72 eV
We can use the formula: λ = h/p.
In this case, we know that p = 7.6 × 10-35 kg∙m/s.
Thus, λ = (6.63 × 10-34) / (7.6 × 10-35) = 8.72 eV
2. The momentum of photon A is pA and the momentum of photon B is pB. If pB = 1/3 pA,
which of the following of the de Broglie wavelength is correct?
λ = 1/3 λ
A
λ = 3λ
A
λ =λ
A
B
B
B
λ = 6λ
A
B
None of the above is correct
We can use the formula: λ = h/p.
pB = 1/3 pA
Thus, λA = 1/3 λB
3. What is the momentum of a photon with a de Broglie wavelength of 5.4 m? Estimate h as
6.63 × 10-34J s.
1.01 × 10 kg∙m/s
-34
1
1.23 × 10 kg∙m/s
-34
1.23 × 10 kg∙m/s
-35
1.45 × 10 kg∙m/s
-34
2.15 × 10 kg∙m/s
-34
We can use the formula: p = h/ λ.
In this case, we know that λ = 5.4 m.
Thus, p = (6.63 × 10-34) /5.4 = 1.23 × 10-34 kg∙m/s
4. If the frequency of a bunch of lights waves is 5.4 × 1015Hz, what’s the energy of each
photon? Estimate h as 4.14 × 10-15 eVs.
18.7 eV
21.4 eV
22.4 eV
23.5 eV
24.5 eV
We can use the formula: E = hf.
In this case, we know that f = 5.4 × 1015Hz.
Thus, E = (4.14 × 10-15) (5.4 × 1015) = 22.4eV
5. If the momentum of a photon is 8.45 × 10-28 kg∙m/s, what is the energy of the photon?
Estimate h as 6.63 × 10-34J s, and c as 3.00 × 108m/s .
1.58eV
1.76eV
2.03eV
2.67eV
3.25eV
2
We can use the formula p = h/λ to find λ.
λ = h/p = (6.63 × 10-34) / (8.45 × 10-28) = 7.85 × 10-7 m
Then, we can use the formula: f = c/λ = (3.0 × 108) / (7.85 × 10-7) = 3.82 × 1014 Hz
Thus, the energy of the photon E = (4.14 × 10-15) (3.82 × 1014) = 1.58eV.
6. A light with a frequency of 8.3 × 1014 Hz shines on calcium, which has a work function of
2.9 eV. What is the maximum kinetic energy of an electron released due to the
photoelectric effect? Estimate h as 4.14 × 10-15 eVs.
0.54 eV
1.6 eV
2.4 eV
4.8eV
There will not be any electrons released
We will use the equation KEmax = hf - ϕ
We plug in our known values and get
KEmax = (8.3 × 1014) (4.14 × 10-15) -2.9 = 0.54eV.
7. In the figure below when U = 0.5 V, the current I just decreases to 0. The frequency of the
light is 2.2 × 1015 Hz. What’s the work function of the metal sheet K? (Estimate h as 4.14 ×
10-15 eVs and electrons have a charge of -1.6 × 10-19C.
8.61eV
9.11eV
0.55eV
9.61eV
8.83eV
the kinetic energy of an electron emitted KEmax = Uq = hf-ϕ.
Thus, ϕ = hf-Uq = 4.14 × 10-15 × 2.2 × 1015-0.5 × 1.6 × 10-19 = 9.108eV-0.5eV = 8.61eV
3