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Fundamentals Of Physics
Found in: Page 1246
Fundamentals Of Physics

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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Short Answer

On which quantum numbers does the energy of an electron depend in (a) a hydrogen atom and

(b) a vanadium atom?

  1. The quantum number that the energy of an electron depends in a hydrogen atom is the principal quantum number n .
  2. The quantum number that the energy of an electron depends in a vanadium atom is the principal quantum number n and l .
See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: The given data

Atoms are Hydrogen ( Z = 1 ) and Vanadium ( Z = 23 ) .

Step 2: Understanding the concept of energy

Using the energy of an electron in the nth Bohr's orbit of an atom, we can calculate the energy of an electron of the required atom that are hydrogen-like species using their atomic number and the principal quantum number of the subshell where the electron is present. But for transition metals that are not hydrogen-like species, the energy of the electron also depends on the azimuthal number irrespective of the dependence of energy on the principal quantum number.

Formula:

The energy of an electron in the nth Bohr's orbit or shell of the atom,

E=-13.6Z2n2 ……….(i)

Step 3: a) Calculation of the energy in the hydrogen atom

According to the concept and equation (i), we can say that the energy of hydrogen or any other hydrogen-like species depends on the principal quantum number n .

Hence, it depends on the principal quantum number n .

Step 4: b) Calculation of energy in vanadium atom

Now, as Vanadium is not hydrogen-like species and is a transition metal, the energy of this atom due to transition depends on both the principal n and azimuthally quantum number l that is, the energy dependence is on the value n+l.

Hence, the energy of this atom depends on n and l .

Most popular questions for Physics Textbooks

How many electron states are in these sub-shells: (a) n = 4,l = 3 ; (b) n = 3, l = 1 ; (c) n = 4, l = 1 ; (d) n = 2, l = 0 ?

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.

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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?

Suppose that a hydrogen atom in its ground state moves 80 cm through and perpendicular to a vertical magnetic field that has a magnetic field gradient dBdz=1.6×102T . (a) What is the magnitude of force exerted by the field gradient on the atom due to the magnetic moment of the atom’s electron, which we take to be Bohr magnetron? (b) What is the vertical displacement of the atom in the 80cm of travel if its speed is 1.2×105 m/s?

An electron in an atom of gold is in a state with n = 4. Which of these values of are possible for it:-3, 0, 2, 3, 4, 5?

Can an incoming intercontinental ballistic missile be destroyed by an intense laser beam? A beam of intensity 108 W/m2would probably burn into and destroy a non-spinning missile in 1s. (a) If the laser had 5.0MW power, 3 μmwavelength, and a beam diameter of (a very powerful laser indeed), would it destroy a missile at a distance of 3000km ? (b) If the wavelength could be changed, what maximum value would work? Use the equation for the central diffraction maximum as given by Eq. 36-12 sin θ=1.22λd.
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