A hydrogen atom absorb a photon of visible light and it's electron leaves from n=4 and enters n=6 energy calculate the change of energy of the atom and wavelength in (nm) of the photon (R=-2.18×10^-18)
To calculate the change in energy of the atom, we can use the formula:
ΔE = E_final - E_initial
Where ΔE is the change in energy, E_final is the final energy level, and E_initial is the initial energy level.
Given that the electron leaves from n = 4 and enters n = 6, we can substitute these values into the formula:
ΔE = (-2.18 × 10^-18) * ((1/n_final^2) - (1/n_initial^2))
ΔE = (-2.18 × 10^-18) * ((1/6^2) - (1/4^2))
ΔE = (-2.18 × 10^-18) * (1/36 - 1/16)
ΔE = (-2.18 × 10^-18) * (16/576 - 36/576)
ΔE = (-2.18 × 10^-18) * (-20/576)
ΔE = 3.817 × 10^-20 J
To calculate the wavelength of the photon, we can use the formula:
ΔE = hc/λ
Where ΔE is the change in energy, h is Planck's constant (6.62607015 × 10^-34 J*s), c is the speed of light (2.998 × 10^8 m/s) and λ is the wavelength.
Substituting the values:
3.817 × 10^-20 = (6.62607015 × 10^-34) * (2.998 × 10^8) / λ
Solving for λ:
λ = (6.62607015 × 10^-34) * (2.998 × 10^8) / (3.817 × 10^-20)
λ = 5.226 × 10^-7 m
Converting this to nanometers:
λ = 5.226 × 10^-7 * 10^9
λ = 522.6 nm
Therefore, the change in energy of the atom is 3.817 × 10^-20 J and the wavelength of the photon is 522.6 nm.
To calculate the change in energy of the hydrogen atom, we can use the formula:
ΔE = (Efinal - Einitial)
where Efinal is the energy of the final state and Einitial is the energy of the initial state.
Given that the electron moves from n = 4 to n = 6, we can find the energies of these states using the formula:
En = -Rhc/n^2
where R is the Rydberg constant (-2.18×10^-18 J), h is the Planck constant (6.626×10^-34 J⋅s), and c is the speed of light (3.00×10^8 m/s).
Calculating the initial energy (Einitial):
Einitial = -Rhc/(4^2)
Calculating the final energy (Efinal):
Efinal = -Rhc/(6^2)
Substituting the given values:
Einitial = (-2.18×10^-18 J)(6.626×10^-34 J⋅s)(3.00×10^8 m/s)/(4^2)
Efinal = (-2.18×10^-18 J)(6.626×10^-34 J⋅s)(3.00×10^8 m/s)/(6^2)
Calculating ΔE:
ΔE = Efinal - Einitial
Finally, to calculate the wavelength of the photon, we can use the relation between energy and wavelength:
ΔE = hc/λ
where λ is the wavelength.
Rearranging the equation:
λ = hc/ΔE
Substituting the given values:
λ = (6.626×10^-34 J⋅s)(3.00×10^8 m/s)/ΔE
Calculating ΔE will give us the change in energy of the atom, and calculating λ will give us the wavelength of the photon.
To calculate the change in energy of the hydrogen atom and the wavelength of the absorbed photon, we can use the formula:
ΔE = -Rh * [(1/n_final^2) - (1/n_initial^2)]
Where:
ΔE is the change in energy of the atom
Rh is the Rydberg constant (in this case, Rh = -2.18 × 10^-18 J)
n_final is the final principal quantum number (in this case, n = 6)
n_initial is the initial principal quantum number (in this case, n = 4)
Let's substitute the given values into the formula to calculate ΔE:
ΔE = -(-2.18 × 10^-18 J) * [(1/6^2) - (1/4^2)]
= -(-2.18 × 10^-18 J) * [1/36 - 1/16]
= -(-2.18 × 10^-18 J) * [16/576 - 36/576]
= -(-2.18 × 10^-18 J) * [-20/576]
= (2.18 × 10^-18 J) * (20/576)
= 0.075279 × 10^-18 J
= 7.5279 × 10^-20 J
So, the change in energy of the hydrogen atom is 7.5279 × 10^-20 J.
To calculate the wavelength of the photon, we can use the equation:
λ = c / ν
Where:
λ is the wavelength of the photon
c is the speed of light in a vacuum (c = 3 × 10^8 m/s)
ν is the frequency of the photon
The frequency can be calculated using the energy of the photon and the Planck's equation:
E = h * ν
Where:
E is the energy of the photon
h is Planck's constant (h = 6.63 × 10^-34 J*s)
Let's find the frequency:
E = h * ν
ν = E / h
= (7.5279 × 10^-20 J) / (6.63 × 10^-34 J*s)
= 1.136 x 10^14 s^-1
Now, let's calculate the wavelength:
λ = c / ν
= (3 × 10^8 m/s) / (1.136 x 10^14 s^-1)
= 2.64 × 10^-7 m
= 264 nm
Therefore, the change in energy of the atom is 7.5279 × 10^-20 J and the wavelength of the absorbed photon is 264 nm.