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.

Please enter the atomic number (Z) of the ion gas in the tube. For example, for hydrogen, please enter '1'.

la energia balistica es igual al cambio de energia de transicion.

qv= kz^2((1/nf^2)-(1/ni^2)))
solo hay una emision del espectro, entonces la energia balistica solo es capaz de exictar del estado n=1 al estado n=2:
qv= kz^2((1/2^2)-(1/1^2)))=3/4*kz^2
z=((1.641e-9*91.9)/(0.75*2.18e-18))^0.5
z=3, el gas es Li, Perdon por colocar la rpta en español, no soy muy bueno escribiendo en ingles

Muchas gracias mi amigo ;)

you are welcome !!

To determine the chemical identity of the ion gas in the tube, we need to understand the concept of ionization energy and the relationship between electron energy levels.

In the given problem, the ion gas consists of ions of the same element, each with one remaining electron. These electrons are accelerated across the potential difference and collide with the gaseous ions. When the voltage is below a threshold value of 91.9V, the emission spectrum is completely blank, indicating that no energy transitions are occurring.

At a plate voltage of 91.9V, the emission spectrum consists of a single line. This suggests that at this voltage, an electron is undergoing an energy transition from an excited state to a lower energy state, emitting a specific wavelength of light. The energy required for this transition to occur is equal to the ionization energy of the ion gas.

Now, the ionization energy is defined as the amount of energy required to remove one electron from an atom or ion in the gas phase. This energy is specific to each element and is related to its atomic number (Z).

To find the chemical identity of the ion gas, we need to find the element with an ionization energy equal to the threshold voltage of 91.9V. Fortunately, there is a relationship between energy and voltage in an electric field. The energy (E) of a charged particle accelerated across a voltage (V) is given by the equation:

E = qV

where q is the charge of the particle. In this case, since the particle is an electron, the charge (q) is -1. Thus, the energy of the electron accelerated across the voltage is -1 * 91.9eV.

We can now use the ionization energy formula:

Ionization Energy = -1 * 91.9eV

By referring to the ionization energy values of different elements, we can determine which element has an ionization energy closest to -1 * 91.9eV. Looking up the ionization energy values, we find that the element with the closest ionization energy is argon (Ar), with an ionization energy of approximately 15.8eV.

The atomic number (Z) of argon is 18.

Therefore, the chemical identity of the ion gas in the tube is argon, and the atomic number (Z) of the ion gas is 18.