a) [tex]5.64\cdot 10^{14} Hz[/tex]
b) [tex]3.74\cdot 10^{-19}J[/tex]
c) [tex]2.25\cdot 10^5 J[/tex]
d) [tex]-2.2\cdot 10^{-18} J[/tex]
e) [tex]-5.4\cdot 10^{-19} J[/tex]
f) [tex]1.66\cdot 10^{-18} J[/tex]
Explanation:
a)
The frequency and the wavelength of a light wave are related by the so-called wave equation:
[tex]c=f\lambda[/tex]
where
c is the speed of light
f is the frequency of light
[tex]\lambda[/tex] is the wavelength
In this problem, we have:
[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light
[tex]\lambda=532 nm = 532\cdot 10^{-9} m[/tex] is the wavelength of the green light emitted by the laser
Solving for f, we find the frequency:
[tex]f=\frac{c}{\lambda}=\frac{3.0\cdot 10^8}{532\cdot 10^{-9}}=5.64\cdot 10^{14} Hz[/tex]
b)
The energy of a photon is given by the equation:
[tex]E=hf[/tex]
where
E is the energy
h is the Planck constant
f is the frequency of the photon
For the photon in this problem we have:
[tex]h=6.63\cdot 10^{-34}Js[/tex] is the Planck's constant
[tex]f=5.64\cdot 10^{14} Hz[/tex] is the frequency of the photon
Substituting, we find the energy of the photon:
[tex]E=(6.63\cdot 10^{-34})(5.64\cdot 10^{14})=3.74\cdot 10^{-19}J[/tex]
c)
1 mole of a substance is the amount of substance that contains a number of particles equal to Avogadro number:
[tex]N_A=6.022\cdot 10^{23}[/tex]
This means that 1 mole of photons will contain a number of photons equal to the Avogadro number.
Here, we know that the energy of a single photon is
[tex]E=3.74\cdot 10^{-19}J[/tex]
Therefore, the energy contained in one mole of photons of this light will be
[tex]E' = N_A E[/tex]
And substituting the two numbers, we get:
[tex]E'=(6.022\cdot 10^{23})(3.74\cdot 10^{-19})=2.25\cdot 10^5 J[/tex]
d)
According to the Bohr's model, the orbital energy of an electron in the nth-level of the atom is
[tex]E_n = -13.6\frac{1}{n^2}[/tex] [eV]
where
n is the level of the orbital
The energy is measured in electronvolts
In this problem, we have an electron in the ground state, so
n = 1
Therefore, its energy is
[tex]E_1=-13.6\frac{1}{1^2}=-13.6 eV[/tex]
And given the conversion factor between electronvolts and Joules,
[tex]1 eV = 1.6\cdot 10^{-19} J[/tex]
The energy in Joules is
[tex]E_1 = -13.6 \cdot (1.6\cdot 10^{-19})=-2.2\cdot 10^{-18} J[/tex]
e)
As before, the orbital energy of an electron in the hydrogen atom is
[tex]E_n = -13.6\frac{1}{n^2}[/tex] [eV]
where:
n is the level of the orbital
The energy is measured in electronvolts
Here we have an electron in the
n = 2 state
So its energy is
[tex]E_2=-13.6\cdot \frac{1}{2^2}=-3.4 eV[/tex]
And converting into Joules,
[tex]E_2=-3.4 (1.6\cdot 10^{-19})=-5.4\cdot 10^{-19} J[/tex]
f)
The energy required for an electron to jump from a certain orbital to a higher orbital is equal to the difference in energy between the two levels, so in this case, the energy the electron needs to jump from the ground state (n=1) to the higher orbital (n=2) is:
[tex]\Delta E = E_2-E_1[/tex]
where:
[tex]E_2=-5.4\cdot 10^{-19} J[/tex] is the energy of orbital n=2
[tex]E_1=-2.2\cdot 10^{-18}J[/tex] is the energy of orbital n=1
Substituting, we find:
[tex]\Delta E=-5.4\cdot 10^{-19}-(-2.2\cdot 10^{-18})=1.66\cdot 10^{-18} J[/tex]
