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Find the energy (in joules) of the photon that is emitted when the electron in a hydrogen atom undergoes a transition from the n = 8 energy level to produce a line in the Paschen series (with principle quantum number of 3).

User Petrelharp
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2 Answers

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Final answer:

To find the energy of the photon emitted by the electron transition from the n = 8 to the n = 3 energy level in a hydrogen atom, we can use the equations E = hc/λ and 1/λ = R(1/n₁² - 1/n₂²), where E is the energy of the photon, h is Planck's constant, c is the speed of light, λ is the wavelength of the photon, and R is the Rydberg constant. By substituting the given values, we can calculate the energy and wavelength of the photon.

Step-by-step explanation:

When an electron in a hydrogen atom undergoes a transition from the n = 8 energy level to produce a line in the Paschen series with a principle quantum number of 3, we can calculate the energy of the photon emitted.

The energy of a photon can be calculated using the equation E = hc/λ, where E is the energy of the photon, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the photon.

The transition from n = 8 to n = 3 corresponds to the emission of a photon with a wavelength in the infrared region of the electromagnetic spectrum. Using the equation E = hc/λ and substituting the values, the energy of the photon can be calculated as follows:

E = (6.63 x 10^-34 J·s) * (3.00 x 10^8 m/s) / λ

To find the value of λ, we can use the formula for the wavelength of a line in the Paschen series:

1/λ = R(1/n₁² - 1/n₂²)

where R is the Rydberg constant (1.097 x 10^7 m⁻¹), n₁ is the principle quantum number of the lower energy level, and n₂ is the principle quantum number of the higher energy level. Substituting the values, we can find the value of λ. Once we have the value of λ, we can substitute it into the equation E = hc/λ to calculate the energy of the photon in joules.

Therefore, the energy (in joules) of the photon that is emitted when the electron in a hydrogen atom undergoes a transition from the n = 8 energy level to produce a line in the Paschen series with a principle quantum number of 3 can be calculated using the equations mentioned above.

User Mensah
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Answer:

The energy is
E = 2.08*10^(-19)J

Step-by-step explanation:

From the question we are told that

The energy level of the photon is
n=8

Generally the spatial frequency(
(1)/(\lambda)) of this transition is mathematically represented as


(1)/(\lambda ) = R * ((1)/(n_1) - (1)/(n) )

Where R is the Rydberg constant with a value of
1.09 * 10^7m^(-1)


n_1 is the principle quantum number for Paschen series with value given a s
n_1 = 3


\lambda is the wavelength of the photon

Now substituting values


(1)/(\lambda ) = 1.097 *10^7 ((1)/(3^2) - (1)/(8^2) )


= 1.047 *10^6m^(-1)

Now the energy of this photon is mathematically represented as


E = h * c * (1)/(\lambda)

where h is the Planck's constant with value
h = 6.626 *10 ^ {-34} J \cdot sec

c is the speed of light with value
c = 3.0*10^8 m/s

Substituting values


E = 6.626 *10^(-34) * 3*10^8 * 1.047 *10^6


= 2.08*10^(-19)J

User SpenserJ
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