a. What is the difference between binding energy and ionization energy? b. An el

Question

a. What is the difference between binding energy and ionization energy? b. An electron is in the second excited orbit of hydrogen, corresponding to n = 3. Find: i. the radius of the orbit ii. the wavelength of the electron in this orbit a. What is the basic process that occurs in the nucleus during normal beta^- decay? Write basic process down the general nuclear reaction equation. b. Tritium^3_1H decays by beta^- with a half-life of 12.33 years and releasing emission 0.0186 MeV of energy at each decay. Write down the nuclear reaction equation. c. Find the radioactive decay constant for tritium. d. A person accidentally receives exposure to tritium of activity 4.6 times 10^3 Bq. If the person has a mass of 80 kg, estimate the absorbed radiation dose in grays (J/kg) in a 24 hour period.

Explanation / Answer

11.a.) Binding energy of a system is the energy released when the constituents of the system are brought from inifinity to form the system.

Binding enery can also be defined as the energy required to separate the constituents of a system to infinite distancs from each other.

The system mentioned here may be a nucleus where the constituents would be the nucleons, i.e., protons and neutrons.

Another example of the system may be the whole atmo, where the constituents are nucleus and electrons.

Ionization enery is just the minimum energy required to ionize an atom. When the electron of an atom is given enough energy so that its energy becomes zero from negative, then the electron is free from the attraction of nucleus and is at infinite distance from nucleus and the atom in this state is said to be ionized.

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11.b.) (i) The radius of a hydrogen like atom for nth orbit is given by

rn = n2a0/Z

where, a0 = Bohr radius = 5.29177x10 -11m

Here, we have Z = 1 for hydrogen atoms and we have to find the radius of the n=3 orbit for Hydrogen. So we have:

r = (n2a0)/Z = [9X(5.29177x10 -11m)]/1 = 47.62x10-11m

  

11.b.) (ii) The energy for nth orbit for Hydrogen atom is given by;

En = -(13.6eV)/n2

So, energy for the n=3 Bohr orbit for Hydrogen is given by:

E3 =  -(13.6eV)/(3)2 = -1.511eV

Now E = hc/,

where E = energy = -1.511eV

h is Planck's constant = 6.63×10-34 m2 kg / s

c = speed of light = 3X108m/s

So using  E = hc/, we get

1.511eV = (6.63×10-34 m2 kg / s)(3X108m/s)/

or (1.511)(1.6X10-19J) = (6.63×10-34 m2 kg / s)(3X108m/s)/

or = 8.22X10-7m = 822nm

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This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification or correction I will be happy to oblige...

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