/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 A hydrogen atom initially in its... [FREE SOLUTION] | 91Ó°ÊÓ

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A hydrogen atom initially in its ground level absorbs a photon, which excites the atom to the \(n=3\) level. Determine the wavelength and frequency of the photon.

Short Answer

Expert verified
To solve this problem, firstly the Rydberg formula has to be applied to determine the wavelength of the photon, followed by applying the speed of light formula to find out the frequency. The final results would depend on exact values of these constants.

Step by step solution

01

Apply the Rydberg formula

The Rydberg formula in terms of wavelength is given by \(\frac{1}{\lambda} = R_H (\frac{1}{n_i^2} - \frac{1}{n_f^2})\) where \(R_H\) is the Rydberg constant, \(n_i\) is the initial energy level and \(n_f\) is the final energy level. In this case, \(n_i = 1\) (ground level) and \(n_f = 3\). Apply these values to the Rydberg formula to find the wavelength.
02

Solve for the wavelength

Applying the values, we obtain \(\frac{1}{\lambda} = R_H (\frac{1}{1^2} - \frac{1}{3^2})\). We solve this equation for \(\lambda\) by taking the reciprocal.
03

Calculate the frequency

Now that we have the wavelength, we can calculate the frequency \(\nu\) using the speed of light formula \(c = \lambda \nu\), where c is the speed of light (approximately \(3*10^8 m/s\)). Solve this equation for \(\nu\).
04

Substitute the values

Substitute the value for \(\lambda\) from Step 2 and the speed of light into the equation \(\nu = c / \lambda\) to find the frequency of the photon.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrogen Atom Energy Levels
The hydrogen atom, with its single electron and nucleus, serves an essential model for understanding quantum mechanics and how particles behave at very small scales. Energy levels in a hydrogen atom are quantized, which means the electron can only exist in certain discrete energy states.

These energy states are determined by Quantum Mechanics and are defined by principle quantum numbers, denoted as \(n\). The ground state, or the lowest energy level, is represented by \(n=1\). When an electron absorbs energy, it moves to a higher energy level, denoted by a larger integer value of \(n\).

How Energy Levels Relate to Photons

When an electron jumps from a lower energy level \(n_i\) to a higher one \(n_f\), it absorbs a photon of light, and conversely, it emits a photon when it falls from a higher energy level to a lower one. The energy of the absorbed or emitted photon corresponds to the difference between the two energy levels. Knowing the initial and final states of the electron allows us to use equations like the Rydberg formula to calculate this energy difference and the corresponding photon characteristics.
Photon Wavelength Calculation
The wavelength of a photon is intrinsically related to its energy. In fact, the wavelength determines the color of light that we see or, in more technical terms, the type of electromagnetic radiation. Using the Rydberg formula, which links the energy levels of the hydrogen atom to the wavelength of emitted or absorbed light, we can calculate the wavelength of the photons associated with electron transitions.

The Rydberg formula for calculating the wavelength of light absorbed or emitted during these transitions is:\[\frac{1}{\text{λ}} = R_H \left(\frac{1}{{n_i}^2} - \frac{1}{{n_f}^2}\right)\]. Here, \(\text{λ}\) is the wavelength, \(R_H\) is the Rydberg constant, \(n_i\) is the initial energy level, and \(n_f\) is the final energy level.

Process of Calculating Wavelength

In the Rydberg formula, we use the values of \(n_i\) and \(n_f\) for the hydrogen atom's electron's initial and final states to determine the inverse of the wavelength. To find the actual wavelength, we take the reciprocal of this result. As students, by performing these calculations, you'll be able to predict not only the energy but also the type of light involved in these tiny atomic events.
Photon Frequency Calculation
Frequency is another way to express the characteristics of a photon and is directly related to its wavelength. The frequency is the number of wave cycles that pass a given point per unit of time, typically measured in Hertz (Hz).

Using the relationship between the speed of light \(c\), wavelength \(\text{λ}\), and frequency \(u\), defined by the formula \(c = \text{λ} u\), we can determine the frequency of a photon once we know its wavelength. To solve for the frequency, we rearrange the formula to \(u = \frac{c}{\text{λ}}\), where \(c\) is the constant speed of light, approximately \(3 \times 10^8\) meters per second.

Converting Wavelength to Frequency

Substituting the calculated wavelength from the previous step into the equation allows us to find the photon's frequency. Understanding the frequency is useful for various applications, such as determining the energy of the photon and exploring its interactions with matter. For instance, in telecommunications, specific photon frequencies correspond to different channels of information transmission.

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Most popular questions from this chapter

(a) Using the Bohr model, calculate the speed of the electron in a hydrogen atom in the \(n=1,2,\) and 3 levels. (b) Calculate the orbital period in each of these levels. (c) The average lifetime of the first excited level of a hydrogen atom is \(1.0 \times 10^{-8} \mathrm{~s}\). In the Bohr model, how many orbits does an electron in the \(n=2\) level complete before returning to the ground level?

Two stars, both of which behave like ideal blackbodies, radiate the same total energy per second. The cooler one has a surface temperature \(T\) and a diameter 3.0 times that of the hotter star. (a) What is the temperature of the hotter star in terms of \(T ?\) (b) What is the ratio of the peak-intensity wavelength of the hot star to the peak-intensity wavelength of the cool star?

A beam of electrons is accelerated from rest through a potential difference of \(0.100 \mathrm{kV}\) and then passes through a thin slit. When viewed far from the slit, the diffracted beam shows its first diffraction minima at \(\pm 14.6^{\circ}\) from the original direction of the beam. (a) Do we need to use relativity formulas? How do you know? (b) How wide is the slit?

Use Balmer's formula to calculate (a) the wavelength, (b) the frequency, and (c) the photon energy for the \(\mathrm{H}_{\gamma}\) line of the Balmer series for hydrogen.

DATA As an amateur astronomer, you are studying the apparent brightness of stars. You know that a star's apparent brightness depends on its distance from the earth and also on the fraction of its radiated energy that is in the visible region of the electromagnetic spectrum. But, as a first step, you search the Internet for information on the surface temperatures and radii of some selected stars so that you can calculate their total radiated power. You find the data given in the table. $$ \begin{array}{l|cccc} \text { Star } & \text { Polaris } & \text { Vega } & \text { Antares } & \alpha \text { Centauri B } \\ \hline \text { Surface temperature (K) } & 6015 & 9602 & 3400 & 5260 \\ \hline \begin{array}{l} \text { Radius relative to that of the } \\ \text { sun }\left(R_{\text {sun }}\right) \end{array} & 46 & 2.73 & 883 & 0.865 \\ & & & & \end{array}$$ The radius is given in units of the radius of the sun, \(R_{\text {sun }}=6.96 \times 10^{8} \mathrm{~m}\) The surface temperature is the effective temperature that gives the measured photon luminosity of the star if the star is assumed to radiate as an ideal blackbody. The photon luminosity is the power emitted in the form of photons. (a) Which star in the table has the greatest radiated power? (b) For which of these stars, if any, is the peak wavelength \(\lambda_{\mathrm{m}}\) in the visible range \((380-750 \mathrm{nm}) ?\) (c) The sun has a total radiated power of \(3.85 \times 10^{26} \mathrm{~W} .\) Which of these stars, if any, have a total radiated power less than that of our sun?

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