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Chapter 38: Problem 49

What is the shortest possible wavelength of the Lyman series in hydrogen?

Short Answer

Expert verified
Answer: The shortest possible wavelength of the Lyman series in hydrogen is approximately 91.1 nm.

Step by step solution

01

Recall the formula for the hydrogen spectral lines

To calculate the wavelength of the spectral lines in hydrogen, we will use the Rydberg formula for hydrogen: $$\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$ In this formula, λ represents the wavelength, 𝑅_H is the Rydberg constant for hydrogen (approximately 1.097 x 10^7 m⁻¹), n₁ and n₂ are the principal quantum numbers of the initial and final energy levels, respectively.
02

Determine n₁ and n₂ for the Lyman series

For the Lyman series, electron transitions occur between the first energy level (n = 1) and any higher energy level (n > 1). To find the shortest wavelength, we need the transition with the highest energy difference, which corresponds to an electron transitioning from n -> infinity (n₂ = ∞) to the first energy level (n₁ = 1).
03

Plug the values into the Rydberg formula

Now that we have n₁ = 1 and n₂ = ∞, we can plug these values into the Rydberg formula: $$\frac{1}{\lambda} = R_H \left(\frac{1}{1^2} - \frac{1}{\infty^2}\right)$$
04

Simplify the equation

Now, simplifying the equation, we get: $$\frac{1}{\lambda} = R_H (1 - 0)$$ $$\frac{1}{\lambda} = R_H$$
05

Find the shortest possible wavelength

Now, we can solve for λ by taking the reciprocal of both sides of the equation: $$\lambda = \frac{1}{R_H}$$ Plugging in the value of 𝑅_H (1.097 x 10^7 m⁻¹): $$\lambda = \frac{1}{1.097 \times 10^7 \mathrm{m^{-1}}}$$ Thus, the shortest possible wavelength of the Lyman series in hydrogen is: $$\lambda ≈ 9.11 \times 10^{-8} \mathrm{m}$$ or approximately 91.1 nm.

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Most popular questions from this chapter

Hund's rule, a component of the Aufbauprinzip (construction principle), states that as one moves across the periodic table, with increasing atomic number, the available electron subshells are filled successively with one electron in each orbital, their spins all parallel; only when all orbitals in a subshell contain one electron are second electrons, with spins opposite to the first, placed in the orbitals. Explain why the ground state electron configurations of successive elements should follow this pattern.

Find the energy difference between the ground state of hydrogen and deuterium (hydrogen with an extra neutron in the nucleus)

A ruby laser consists mostly of alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) and a small amount of chromium ions, responsible for its red color. One such laser of power \(3.00 \mathrm{~kW}\) emits light pulse of duration \(10.0 \mathrm{~ns}\) and of wavelength \(685 \mathrm{nm}\). a) What is the energy of the photons in the pulse? b) Determine the number of chromium atoms undergoing stimulated emission to produce this pulse.

A muon is a particle very similar to an electron. It has the same charge but its mass is \(1.88 \cdot 10^{-28} \mathrm{~kg}\). a) Calculate the reduced mass for a hydrogen-like muonic atom consisting of a single proton and a muon. b) Calculate the ionization energy for such an atom, assuming the muon starts off in its ground state.

Prove that the period of rotation of an electron on the \(n\) th Bohr orbit is given by: \(T=n^{3} /\left(2 c R_{\mathrm{H}}\right),\) with \(n=1,2,3, \ldots\)
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