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Chapter 19: Problem 87

An electron in hydrogen atom first jumps from second excited state to first excited state and then from first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons emitted in these two cases be \(a, b\) and \(c\) respectively, then (A) \(c=\frac{1}{a}\) (B) \(a=\frac{9}{4}\) (C) \(b=\frac{5}{27}\) (D) \(c=\frac{5}{27}\)

Short Answer

Expert verified
The correct ratios we found for the problem are: \(a = \frac{5}{3}\), \(b = \frac{3}{5}\), and \(c = \frac{3}{5}\). None of the given options match these calculated ratios.

Step by step solution

01

Identify the relevant equations

Here, we will use the Rydberg formula for calculating the energy difference, and relationships between energy, momentum and wavelength of emitted photons. 1) Rydberg formula: \(\frac{1}{\lambda} = R_H \left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)\) where \(\lambda\) is the wavelength of emitted photon, \(R_H\) is Rydberg constant, and \(n_i\) and \(n_f\) are the initial and final energy levels for electron. 2) Planck's equation: \(E = h\nu = \frac{hc}{\lambda}\) where \(E\) is the energy of the photon, \(h\) is Planck's constant, \(\nu\) is the frequency of the photon, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the photon. 3) Momentum of photon: \(p = \frac{h}{\lambda}\)
02

Find the wavelengths of emitted photons

Using the Rydberg formula, we will find the wavelengths of the two emitted photons for each transition. For electron jumping from second excited state (n=3) to first excited state (n=2): \(\frac{1}{\lambda_1} = R_H \left(\frac{1}{2^2} - \frac{1}{3^2}\right)\) For electron jumping from first excited state (n=2) to ground state (n=1): \(\frac{1}{\lambda_2} = R_H \left(\frac{1}{1^2} - \frac{1}{2^2}\right)\) Now calculate the ratio of the two wavelengths: \(a = \frac{\lambda_1}{\lambda_2} = \frac{9-4}{4-1} = \frac{5}{3}\)
03

Find the momenta of emitted photons

We will now use the momentum of photon formula to find two momenta. Momentum for photon 1: \(p_1 = \frac{h}{\lambda_1}\) Momentum for photon 2: \(p_2 = \frac{h}{\lambda_2}\) Now calculate the ratio of the two momenta: \(b = \frac{p_1}{p_2} = \frac{\lambda_2}{\lambda_1} = \frac{3}{5}\)
04

Find the energies of emitted photons

We will be using Planck's equation to find the two energies. Energy for photon 1: \(E_1 = \frac{hc}{\lambda_1}\) Energy for photon 2: \(E_2 = \frac{hc}{\lambda_2}\) Now calculate the ratio of the two energies: \(c = \frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1} = \frac{3}{5}\)
05

Compare the results with the given options

Now that we have found the ratios \(a, b\) and \(c\), let's compare them to the provided options: (A) \(c=\frac{1}{a}\) ---> Incorrect, as c = 3/5 and a = 5/3 (B) \(a=\frac{9}{4}\) ---> Incorrect, as we found \(a = \frac{5}{3}\) (C) \(b=\frac{5}{27}\) ---> Incorrect, as we found \(b = \frac{3}{5}\) (D) \(c=\frac{5}{27}\) ---> Incorrect, as we found \(c = \frac{3}{5}\) None of the given options match our calculated ratios.

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Key Concepts

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

Rydberg Formula
The Rydberg formula is fundamental in atomic physics for determining the wavelengths of spectral lines of chemical elements. It is particularly important in the study of hydrogen spectral lines, which are prominent tasks in exercises like the JEE MAIN Physics examination.

The formula is given by:
\[\begin{equation}\frac{1}{\lambda} = R_H \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right)\end{equation}\]where \( \lambda \) is the wavelength of the photon emitted during an electron transition, \( R_H \) is the Rydberg constant specific to hydrogen, and \( n_i \) and \( n_f \) represent the initial and final energy levels of the electron, respectively.

Understanding the relationship between energy levels and the emitted photon's wavelength helps students analyze atomic transitions and spectral lines.
Planck's Equation
Max Planck's discovery of the quantization of energy revolutionized physics, leading to the development of quantum mechanics. Planck's equation describes the energy of a photon, the quantum of electromagnetic radiation, and is essential for understanding the photoelectric effect and other phenomena.

Planck's equation can be expressed as:\[\begin{equation}E = hu = \frac{hc}{\lambda}\end{equation}\]where \(E\) is the energy of the photon, \(h\) is Planck's constant (a fundamental constant in physics), \(u\) is the photon's frequency, \(c\) is the speed of light, and \(\lambda\) is the photon's wavelength.

The insight here is that the energy is inversely proportional to the wavelength; thus, shorter wavelengths correspond to more energetic photons. For students dealing with exercises on electromagnetic radiation, knowing this relationship is crucial.
Photon Momentum and Energy
Photons, though massless, carry momentum, a notion that can seem counterintuitive. However, the momentum of a photon is linked to its wave characteristics through the following relation:\[\begin{equation}p = \frac{h}{\lambda}\end{equation}\]Here, \(p\) represents the momentum of the photon, \(h\) is Planck's constant, and \(\lambda\) is the wavelength of the photon.

Moreover, the energy of a photon, given by Planck's equation, and its momentum are directly related by the formula:\[\begin{equation}E = pc\end{equation}\]where \(E\) is energy, \(p\) is momentum, and \(c\) is the speed of light. This relationship becomes key when evaluating the energy and momentum of photons emitted during electronic transitions in atoms.

By mastering these concepts, students will be able to tackle more complex problems involving the interaction of light and matter, such as those found in modern physics and quantum mechanics topics in the JEE MAIN Physics syllabus.

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Most popular questions from this chapter

A hydrogen atom is in an excited state of principle quantum number \(n\). It emits a photon of wavelength \(\lambda\) when returns to the ground state. The value of \(n\) is \((R=\) Rydberg constant \()\) (A) \(\sqrt{\lambda R(\lambda R-1)}\) (B) \(\sqrt{\frac{(\lambda R-1)}{\lambda R}}\) (C) \(\sqrt{\frac{\lambda R}{\lambda R-1}}\) (D) \(\sqrt{\lambda(R-1)}\)

If the potential difference of Coolidge tube producing \(x\)-ray is increased, then choose the correct option(s). (A) the interval between \(\lambda_{k \alpha}\) and \(\lambda_{k \beta}\) increases (B) the interval between \(\lambda_{k \alpha}\) and \(\lambda_{0}\) increases (C) the interval between \(\lambda_{k \beta}\) and \(\lambda_{0}\) increases (D) \(\lambda_{0}\) does not change

When photons of energy \(5 \mathrm{eV}\) strike the surface of a metal \(A\), the ejected photoelectrons have maximum kinetic energy \(K_{A} \mathrm{eV}\) and de Broglie wavelength \(\lambda_{A}\). The maximum kinetic energy of photoelectrons liberated from another metal \(B\) by photons of energy \(5.30 \mathrm{eV}\) is \(K_{B}=\left(K_{A}-1.5\right) . \mathrm{eV}\). If the de Broglie wavelength of these photoelectrons is \(\lambda_{B}=2 \lambda_{A}\), then find \(K_{A}\) and \(K_{B}\)

In a photoelectric effect experiment (A) on increasing intensity and keeping frequency fixed the saturation current decreases. (B) on increasing intensity and keeping frequency fixed the saturation current remains constant. (C) on increasing intensity, saturation current may increase. (D) on increasing frequency saturation current may increase.

If \(N_{0}\) is the original mass of the substance of half-life period \(t_{1 / 2}=5\) years, then the amount of substance left after 15 years is (A) \(N_{0} / 8\) (B) \(N_{0} / 16\) (C) \(N_{0} / 2\) (D) \(N_{0} / 4\)
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