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Chapter 27: Problem 41

Find the wavelength of the radiation emitted when a hydrogen atom makes a transition from the \(n=6\) to the \(n=3\) state. $(\text {tutorial: hydrogen atom })$

Short Answer

Expert verified
Answer: The wavelength of the radiation emitted when a hydrogen atom transitions from the n=6 state to the n=3 state is approximately \(1.10 \times 10^{-6}\) meters.

Step by step solution

01

Recall the Rydberg formula for Hydrogen

The Rydberg formula for a hydrogen atom is given by: \( \frac{1}{\lambda} = R_H(\frac{1}{n_1^2} - \frac{1}{n_2^2}) \) where, \(\lambda\) is the wavelength of the emitted radiation, \(R_H\) is the Rydberg constant for hydrogen, approximately 1.097 x \(10^{7}\) m\(^{-1}\), \(n_1\) is the initial energy level and \(n_2\) is the final energy level. For this problem, the hydrogen atom transitions from the \(n=6\) state to the \(n=3\) state.
02

Plug in the values into the Rydberg formula

Now, we plug the values into the Rydberg formula: \( \frac{1}{\lambda} = 1.097 \times 10^7(\frac{1}{3^2} - \frac{1}{6^2}) \)
03

Evaluate the expression

Simplify the expression and calculate the value of \(\frac{1}{\lambda}\): \( \frac{1}{\lambda} = 1.097 \times 10^7(\frac{1}{9} - \frac{1}{36}) \) \( \frac{1}{\lambda} = 1.097 \times 10^7(\frac{4-1}{36}) \) \( \frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{3}{36} \) \( \frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{1}{12} \)
04

Calculate the wavelength

Finally, to find the wavelength of the emitted radiation, take the reciprocal of the expression we obtained for \(\frac{1}{\lambda}\): \( \lambda = \frac{1}{1.097 \times 10^7 \times \frac{1}{12}} \) Use a calculator to obtain the value of \(\lambda\) as: \(\lambda \approx 1.10 \times 10^{-6} \thinspace m\) The wavelength of the radiation emitted when a hydrogen atom makes a transition from the \(n=6\) to the \(n=3\) state is approximately \(1.10 \times 10^{-6}\) meters.

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Most popular questions from this chapter

Photons of wavelength \(350 \mathrm{nm}\) are incident on a metal plate in a photocell and electrons are ejected. A stopping potential of \(1.10 \mathrm{V}\) is able to just prevent any of the ejected electrons from reaching the opposite electrode. What is the maximum wavelength of photons that will eject electrons from this metal?
What is the cutoff frequency for an x-ray tube operating at \(46 \mathrm{kV} ?\)
Find the Bohr radius of doubly ionized lithium (Li \(^{2+}\) ).
X-rays of wavelength \(10.0 \mathrm{pm}\) are incident on a target. Find the wavelengths of the \(x\) -rays scattered at (a) \(45.0^{\circ}\) and (b) \(90.0^{\circ} .\) (tutorial: Compton scattering)
If the shortest wavelength produced by an x-ray tube is \(0.46 \mathrm{nm},\) what is the voltage applied to the tube?
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