Physics: 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).

Find the energy (in joules) of the photon that is

Ответ:

amandasantiago2001

22.01.2022

Physics

The energy is $E = 2.08*10^{-19}J$

Explanation:

From the question we are told that

The energy level of the photon is n=8

Generally the spatial frequency( $\frac{1}{\lambda}$ ) of this transition is mathematically represented as

$\frac{1}{\lambda } = R * (\frac{1}{n_1} - \frac{1}{n} )$

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

$\frac{1}{\lambda } = 1.097 *10^7 (\frac{1}{3^2} - \frac{1}{8^2} )$

$= 1.047 *10^6m^{-1}$

Now the energy of this photon is mathematically represented as

$E = h * c * \frac{1}{\lambda}$

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

$E = 6.626 *10^{-34} * 3*10^8 * 1.047 *10^6$

$= 2.08*10^{-19}J$

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Ответ:

ausemkattom3034

11.09.2020

Physics

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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).

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