Math Problem Statement

Calculate the wavelength frequency and energy on transition from n =6 in the paschen series

Solution

In the Paschen series of the hydrogen atom, transitions occur when electrons move to the n=3n = 3 energy level from higher energy levels. For a transition from n=6n = 6 to n=3n = 3, we can calculate the wavelength, frequency, and energy using the following steps:

1. Wavelength Calculation (using the Rydberg formula):

The wavelength λ\lambda for a transition between two energy levels in hydrogen can be calculated using the Rydberg formula:

1λ=RH(1n221n12)\frac{1}{\lambda} = R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right)

Where:

  • RHR_H is the Rydberg constant =1.097×107m1= 1.097 \times 10^7 \, \text{m}^{-1}
  • n1=6n_1 = 6 (initial energy level)
  • n2=3n_2 = 3 (final energy level)

Let's calculate λ\lambda:

1λ=1.097×107(132162)\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{3^2} - \frac{1}{6^2} \right)

2. Frequency Calculation:

Once we have the wavelength, the frequency ν\nu can be found using the speed of light equation:

ν=cλ\nu = \frac{c}{\lambda}

Where:

  • c=3.00×108m/sc = 3.00 \times 10^8 \, \text{m/s} (speed of light)

3. Energy Calculation:

The energy EE of the photon can be calculated using the Planck's equation:

E=hνE = h \nu

Where:

  • h=6.626×1034J\cdotpsh = 6.626 \times 10^{-34} \, \text{J·s} (Planck’s constant)

Now, let's perform the calculations.

Step 1: Calculate Wavelength λ\lambda

The wavelength λ\lambda for the transition from n=6n = 6 to n=3n = 3 in the Paschen series is approximately 1.09×106m1.09 \times 10^{-6} \, \text{m} (or 1093.89 nm), which lies in the infrared region of the electromagnetic spectrum.

Now let's calculate the frequency ν\nu and the energy EE of the photon.

Step 2: Calculate Frequency ν\nuThe frequency ν\nu of the photon for this transition is approximately 2.74×1014Hz2.74 \times 10^{14} \, \text{Hz}.

Step 3: Calculate Energy EEThe energy EE of the photon for the transition from n=6n = 6 to n=3n = 3 is approximately 1.82×1019J1.82 \times 10^{-19} \, \text{J}.

Summary:

  • Wavelength: 1.09×106m1.09 \times 10^{-6} \, \text{m} (1093.89 nm)
  • Frequency: 2.74×1014Hz2.74 \times 10^{14} \, \text{Hz}
  • Energy: 1.82×1019J1.82 \times 10^{-19} \, \text{J}

Would you like more details on any part of this? Here are some related questions:

  1. How does the energy of a photon relate to its wavelength?
  2. Can the Paschen series transitions be observed in visible light?
  3. What is the difference between the Paschen and Balmer series?
  4. How is the Rydberg constant derived?
  5. Why do energy levels get closer as nn increases?

Tip: The shorter the wavelength, the higher the energy of the photon.

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Math Problem Analysis

Mathematical Concepts

Quantum Mechanics
Hydrogen Atom Transitions
Electromagnetic Spectrum

Formulas

Rydberg Formula: 1/λ = R_H * (1/n_2^2 - 1/n_1^2)
Frequency Equation: ν = c/λ
Energy Equation: E = hν

Theorems

Bohr Model of the Hydrogen Atom
Planck's Law

Suitable Grade Level

Grades 11-12