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Chemistry, 13.04.2020 11:27 jrfranckowiak

1) Green laser pointers are becoming popular for presentations. The green light is monochromatic
and has a wavelength of 532 nm.
a) What is the frequency of this light?
b) What is the energy in Joules of one photon of this light?
c) What is the energy in Joules of one mole of photons of this light?
d) What is the orbital energy in Joules of an electron in the ground state (n=1) in a hydrogen atom
according to the Bohr model?
e) What is the orbital energy in Joules of an electron in the n=2 state in a hydrogen atom according to
the Bohr model?

1) Green laser pointers are becoming popular for presentations. The green light is monochromatic and

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Answer from: tykiabrown8111

a) $5.64\cdot 10^{14} Hz$

b) $3.74\cdot 10^{-19}J$

c) $2.25\cdot 10^5 J$

d) $-2.2\cdot 10^{-18} J$

e) $-5.4\cdot 10^{-19} J$

f) $1.66\cdot 10^{-18} J$

Explanation:

a)

The frequency and the wavelength of a light wave are related by the so-called wave equation:

$c=f\lambda$

where

c is the speed of light

f is the frequency of light

$\lambda$ is the wavelength

In this problem, we have:

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\lambda=532 nm = 532\cdot 10^{-9} m$ is the wavelength of the green light emitted by the laser

Solving for f, we find the frequency:

$f=\frac{c}{\lambda}=\frac{3.0\cdot 10^8}{532\cdot 10^{-9}}=5.64\cdot 10^{14} Hz$

b)

The energy of a photon is given by the equation:

E=hf

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:

$h=6.63\cdot 10^{-34}Js$ is the Planck's constant

$f=5.64\cdot 10^{14} Hz$ is the frequency of the photon

Substituting, we find the energy of the photon:

$E=(6.63\cdot 10^{-34})(5.64\cdot 10^{14})=3.74\cdot 10^{-19}J$

c)

1 mole of a substance is the amount of substance that contains a number of particles equal to Avogadro number:

$N_A=6.022\cdot 10^{23}$

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

$E=3.74\cdot 10^{-19}J$

Therefore, the energy contained in one mole of photons of this light will be

E' = N_A E

And substituting the two numbers, we get:

$E'=(6.022\cdot 10^{23})(3.74\cdot 10^{-19})=2.25\cdot 10^5 J$

d)

According to the Bohr's model, the orbital energy of an electron in the nth-level of the atom is

$E_n = -13.6\frac{1}{n^2}$ [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

$E_1=-13.6\frac{1}{1^2}=-13.6 eV$

And given the conversion factor between electronvolts and Joules,

$1 eV = 1.6\cdot 10^{-19} J$

The energy in Joules is

$E_1 = -13.6 \cdot (1.6\cdot 10^{-19})=-2.2\cdot 10^{-18} J$

e)

As before, the orbital energy of an electron in the hydrogen atom is

$E_n = -13.6\frac{1}{n^2}$ [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

$E_2=-13.6\cdot \frac{1}{2^2}=-3.4 eV$

And converting into Joules,

$E_2=-3.4 (1.6\cdot 10^{-19})=-5.4\cdot 10^{-19} J$

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:

$\Delta E = E_2-E_1$

where:

$E_2=-5.4\cdot 10^{-19} J$ is the energy of orbital n=2

$E_1=-2.2\cdot 10^{-18}J$ is the energy of orbital n=1

Substituting, we find:

$\Delta E=-5.4\cdot 10^{-19}-(-2.2\cdot 10^{-18})=1.66\cdot 10^{-18} J$

ansver

Answer from: Quest

i think it's c. i'm not sure but i think so.

ansver

Answer from: Quest

i have no clue hope you find the answer

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