A gas in a discharge tube consists entirely of ions of the same element, of the same charge. Each ion has only one remaining electron. The voltage between the electrodes of the gas discharge tube is increased from zero. Electrons are accelerated across the potential, and then hit the gaseous ions within the tube. These are called ballistic electrons. Below a threshold value of 91.9V the emission spectrum is completely blank. At a plate voltage of 91.9V the emission spectrum consists of a single line. Determine the chemical identity of the ion gas in the tube.
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To determine the chemical identity of the ion gas in the tube, we need to refer to the concept of ionization energy.
Ionization energy is the minimum amount of energy required to remove an electron from a neutral atom or molecule in the gas phase. In this case, we know that each ion has only one remaining electron, meaning it has been ionized by the previous collisions.
The fact that the emission spectrum is completely blank below a threshold value of 91.9V suggests that the electrons do not have enough energy to induce any excitation and subsequent emission of photons from the gas ions.
At a plate voltage of 91.9V, however, the emission spectrum consists of a single line, implying that the electrons are now gaining enough energy to excite the gas ions and cause them to emit photons.
The voltage required to produce this emission line indicates the ionization energy of the gas ion. In this case, since the emission line occurs at 91.9V, we can infer that this value is equal to the ionization energy of the gas ion.
To determine the chemical identity, we need to look for an element with an ionization energy of 91.9V. However, voltage is not a direct measure of ionization energy. We need to convert the voltage into a more appropriate unit for ionization energy.
The unit commonly used for ionization energy is electron volts (eV). Since 1V is equal to 1 joule per coulomb, and 1 electron volt is equal to 1.6 x 10^-19 joules, we can convert the 91.9V into electron volts:
91.9V x (1.6 x 10^-19 J/eV) = 1.47 x 10^-17 J
To convert this energy into electron volts, we need to divide by 1.6 x 10^-19 J:
(1.47 x 10^-17 J) / (1.6 x 10^-19 J/eV) = 91.9 eV
Thus, the ionization energy of the gas ion is 91.9 eV.
Now, we need to look for an element with an ionization energy close to 91.9 eV. Referring to the periodic table, we find that the ionization energy of hydrogen, H, is approximately 13.6 eV. However, this does not match the value we obtained.
To find the required ionization energy, we need to multiply the energy of hydrogen by the square of the atomic number (Z), as suggested by Bohr's model of the hydrogen atom. The square of Z for hydrogen is 1^2 = 1.
Therefore, the required ionization energy = 13.6 eV x 1^2 = 13.6 eV.
Since the ionization energy we calculated earlier (91.9 eV) is significantly higher than the required value (13.6 eV), we can rule out hydrogen as the gas ion.
To determine the chemical identity of the ion gas, we need to look for an element with an ionization energy close to 91.9 eV. At this point, I'm unable to provide a specific element that matches this ionization energy value. It is possible that this value corresponds to a hypothetical gas or a gas with unusual properties. Further analysis or experimental data may be needed to identify the chemical identity of the gas ion in the tube.