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Fundamentals Of Physics
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Short Answer

For a helium atom in its ground state, what are quantum numbers (n,l,mI,ms) for the (a) spin-up electron and (b) spin-down electron?

  1. The quantum numbers for the spin-up electron are 1,0,0+12 .
  2. The quantum numbers for the spin-down electron are 1,0,0-12.
See the step by step solution

Step by Step Solution

Step 1: The given data:

The helium atom is its ground state.

Step 2: Understanding the concept of the quantum number of an atom: 

The set of numbers used to describe the position and power of an electron atom is called quantum numbers. There are four quantum numbers, namely, prime numbers, azimuthal, magnetic numbers, and spin quantum. The stored values of the quantum system are given by quantum numbers.

For every value of the principal quantum number, there are n values of l ranging from 0 to n-1 . For every value of the orbital quantum number, there exist 2l+1 values of magnetic quantum number ranging from -l to +l . Thus, the maximum value of this magnetic quantum number is equal to the value of the orbital quantum number. Now, every electron has two spin orientations, thus the value of the magnetic quantum number is determined by these spins.

Step 3: (a) Calculation for the quantum numbers of the spin-up electron

When the helium atom is in the ground state, both its electrons are in 1s state. Thus for each electron with quantum numbers n = 1 , l = 0 and ml=0 ; one of the electrons is in a spin-up quantum state ms=+12 and the other electron is in a spin-down ms=-12 quantum state.

Hence, the quantum numbers for the spin-up electron are 1,0,0+12.

Step 4: (b) Calculation for the quantum numbers of the spin-down electron

Using the concept of calculations in above part (a), the quantum numbers for the spin-down electron is 1,0,0-12.

Most popular questions for Physics Textbooks

Comet stimulated emission. When a comet approaches the Sun, the increased warmth evaporates water from the ice on the surface of the comet nucleus, producing a thin atmosphere of water vapor around the nucleus. Sunlight can then dissociate H2O molecules in the vapor to H atoms and OH molecules. The sunlight can also excite the OH molecules to higher energy levels.

When the comet is still relatively far from the Sun, the sunlight causes equal excitation to the E2 and E1 levels (Fig. 40-28a). Hence, there is no population inversion between the two levels. However, as the comet approaches the Sun, the excitation to the E1 level decreases and population inversion occurs. The reason has to do with one of the many wavelengths—said to be Fraunhofer lines—that are missing in sunlight because, as the light travels outward through the Sun’s atmosphere, those particular wavelengths are absorbed by the atmosphere.

As a comet approaches the Sun, the Doppler Effect due to the comet’s speed relative to the Sun shifts the Fraunhofer lines in wavelength, apparently overlapping one of them with the wavelength required for excitation to the E1 level in OH molecules. Population inversion then occurs in those molecules, and they radiate stimulated emission (Fig. 40 28b). For example, as comet Kouhoutek approached the Sun in December 1973 and January 1974, it radiated stimulated emission at about during mid-January. (a) What was the energy difference E2-E1 for that emission? (b) In what region of the electromagnetic spectrum was the emission?

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