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Mrac [35]
3 years ago
8

What is the longest wavelength of electromagnetic radiation capable of ionizing a ground state hydrogen atom? report your answer

in nanometers?
Health
2 answers:
rjkz [21]3 years ago
8 0
<span>Answer: We can use Rydberg's formula to find the wavelength if we know the energy levels. The formula is: 1/w = R(1/L² - 1/U²) where w is the wavelength, L is the lower energy level, U the upper energy level, and R is Rydberg's constant, experimentally found to be 10,967,758 waves per meter for hydrogen. If the atom is in the n=6 state, that means the lower energy level (L) is 6, and if we ionize it, we remove the electron completely, so the value of U approaches infinity so 1/U becomes zero. Now we can substitute known values into the formula and we obtain: 1/w = 10967758(1/36 - 0) = (10967758)(0.02777777) = 304659. So if 1/w = 304659 waves per meter, w = 3.282 * 10**-6 meter, or 3.282 microns. That is the maximum wavelength of light that will ionize this atom, and it is in the infrared region.</span>
MrRissso [65]3 years ago
6 0

The longest wavelength of electromagnetic radiation to ionize an electron of ground state hydrogen atom is \boxed{{\text{91}}{\text{.2 nm}}}.

Further explanation:

Concept:

Ionization energy is defined as the amount of energy required to remove an electron from an isolated gaseous atom. The energy needed to remove an electron from an atom depends on the position of the electron in the atom. To ionize the electron from ground state of hydrogen, it must be excited to infinity.

According to the Rydberg equation, the energy required toionize the electron from ground state of hydrogen is as follows:

\Delta {\text{E}}=\left( {{{\text{R}}_{\text{H}}}} \right)\left({\frac{1}{{{{\left({{{\text{n}}_{\text{i}}}}\right)}^2}}}-\frac{1}{{{{\left( {{{\text{n}}_{\text{f}}}}\right)}^2}}}}\right)                   …… (1)

Here,{{\text{R}}_{\text{H}} is the Rydberg constant that has the value2.18\times {10^{-18}}{\text{ J}}, {{\text{n}}_{\text{i}} is the quantum number of initial energy level, and {{\text{n}}_{\text{f}} is the quantum number of a final energy level of transition.

The energy of the transition is related to the wavelength as follows:

\lambda=\frac{{hc}}{{\Delta{\text{E}}}}                                 …… (2)

Here, \Delta {\text{E}}is a value of energy of transition, h is the Planck constant and has value 6.626 \times {10^{ - 34}}{\text{ Js}}, \lambdais the wavelength of the corresponding transition, and c is the speed of light and has a value3 \times {10^8}{\text{ m}}{{\text{s}}^{ - 1}}.

Solution:

The ionization of electron from the ground state of hydrogen atom must be excited from n=1 to n=\infty. Therefore, first we have to calculate the energy required for this ionization.

Substitute 2.18\times{10^{-18}}{\text{ J}} for {{\text{R}}_{\text{H}}} and 1 for {{\text{n}}_{\text{i}}}, and \inftyfor {{\text{n}}_{\text{f}}} in equation (1).

\begin{aligned}\Delta{\text{E}}&=\left( {{{\text{R}}_{\text{H}}}} \right)\left({\frac{1}{{{{\left( {{{\text{n}}_{\text{i}}}} \right)}^2}}}-\frac{1}{{{{\left({{{\text{n}}_{\text{f}}}} \right)}^2}}}}\right)\\&=\left({2.18\times {{10}^{-18}}{\text{ J}}} \right)\left({\frac{1}{{{{\left( {\text{1}}\right)}^2}}}-\frac{1}{{{{\left(\infty\right)}^2}}}}\right)\\&={\mathbf{2}}{\mathbf{.18\times1}}{{\mathbf{0}}^{{\mathbf{-18}}}}{\mathbf{J}}\\\end{aligned}

Substitute 2.18\times{10^{-18}}{\text{ J}}for \Delta{\text{E}}, 6.626 \times {10^{ - 34}}{\text{ Js}} for <em>h</em>, and 3 \times {10^8}{\text{ m}}{{\text{s}}^{ - 1}} for c in equation (2).

\begin{aligned}\lambda&=\frac{{hc}}{{\Delta{\text{E}}}}\\&=\frac{{\left( {6.626\times{{10}^{-34}}{\text{ J}} \cdot{\text{s}}}\right)\left({3 \times {{10}^8}{\text{ m}}\cdot{{\text{s}}^{-1}}} \right)}}{{2.18\times{{10}^{-18}}{\text{ J}}}}\\&=\left({9.12\times{{10}^{-8}}{\text{ m}}}\right)\left({\frac{{{{10}^9}{\text{ nm}}}}{{1{\text{ m}}}}}\right)\\&={\mathbf{91}}{\mathbf{.2 nm}}\\\end{aligned}

Hence, the longest wavelength of electromagnetic radiation to ionize an electron of ground state hydrogen atom is 91.2 nm.

Learn more:

1. Ranking of elements according to their first ionization energy:brainly.com/question/1550767

2. Chemical equation representing the first ionization energy for lithium: brainly.com/question/5880605

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Atomic structure

Keywords: transition, hydrogen atom, energy difference, transition from n=1 to infinity, ground state of hydrogen atom, longest wavelength and 91.2 nm.

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