Question

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

Answer 1

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:

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