Physics, 29.07.2019 21:30 bbyitskeke7160

Photons of light with an energy of 656 nm are being directed through a chamber of hydrogen gas atoms. the gas atoms began in their ground state (ni=1). what happens?

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Since there is not quite enough energy for transition from n=1 to n=2. So, the photons will pass through the gas chamber without affecting the gas.

Further explanation:

If the energy of photons is enough to excite all the electrons to go in higher state, the electron absorbs the energy of photon and go from a lower energy state to higher energy state. Transition of electron from lower energy state t higher energy state depends on the energy of photons.

Given:

The wavelength of the photons directed through a chamber of hydrogen gas atoms is $656\text{ nm}$ .

The electron is excited from the energy state n=1 to higher energy state.

Concept:

The energy associated with a photon is given by the following relation.

$E = \dfrac{{hc}}{\lambda}$

Here, is the energy associated with photon, is the plank constant, is the speed of the light and $\lambda$ is the wavelength associated with photon.

Substitute $6.625\times{10^{-34}}\text{ J}\cdot\text{s}$ for , $3.00\times{10^8}\text{ m/s}}$ for and $656\text{ nm}$ for $\lambda$ in the above expression.

$\begin{aligned}E&=\frac{{\left( {6.625 \times {{10}^{ - 34}}\,{\text{Js}}} \right)\left( {3.0 \times {{10}^8}\,{\text{m/s}}} \right)}}{{\left( {656 \times {{10}^{ - 9}}\,{\text{m}}} \right)}}\\&=3.0281 \times {10^{-19}}\,{\text{J}}\\\end{aligned}$

Rydberg equations is used to find out the energy level of an electron while transition from one energy state to another energy state.

$\dfrac{1}{\lambda}=R{z^2}\left({\dfrac{1}{{n_1^2}}-\dfrac{1}{{n_2^2}}}\right)$

Here, $\lambda$ is the wavelength of the photons directed through a chamber, is the Rydberg constant, is the atomic number of hydrogen atom, ${n_1}$ is the lower energy state and ${n_2}$ is the higher energy state.

Substitute $656\times{10^{-9}}\,{\text{m}}$ for $\lambda$ , $1.0973 \times {10^7}\,{{\text{m}}^{{\text{-1}}}}$ for , for and for ${n_1}$ in the above expression.

$\dfrac{1}{{\left( {656 \times {{10}^{ - 9}}\,{\text{m}}} \right)}} = \left( {1.0973 \times {{10}^7}{\mkern 1mu} {{\text{m}}^{{\text{-1}}}}} \right){\left( 1 \right)^2}\left( {\dfrac{1}{{{{\left( 1 \right)}^2}}}-\dfrac{1}{{n_2^2}}}\right)$

Simplify the above expression for ${n_2}$

$\begin{aligned}{n_2}&=\sqrt{\dfrac{1}{{0.861}}}\\&=1.08\\\end{aligned}$

From the above result it can be concluded that photons with wavelength $656\text{ nm}$ does not have enough energy to cause an electron to jump even one level higher.

Thus, there is not quite enough energy for transition from n=1 to n=2. So, the photons will pass through the gas chamber without affecting the gas.

Learn more:

1. A Na+ ion moves from inside a cell link

2. The kinetic energy of the emitted electrons when cesium is exposed to UV rays link

3. The frequency of the light for wavelength 7.1x10^2 nm link

Answer Details:

Grade: College

Subject: Physics

Chapter: Modern physics

Keywords:

Photons, light, energy, 656 nm, 656 times 10^-9 m, 6.56 times 10^-7 m, directed, chamber, hydrogen, gas atom, ground state, ni=1, 1.071, few, absorbed, quite enough, electrons, nf=2.

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