The only visible spectral lines of hydrogen are four Balmer series lines noted at the beginning of Section 7.3. We wish to cause hydrogen gas to glow with its characteristic visible colors.

(a) To how high an energy level must the electrons be exited?

(b) Energy is absorbed in collisions with other particles. Assume that after absorbing energy in one collision, an electron jumps down through lower levels so rapidly that it is in the ground state before another collision occurs. If an electron is to be raised to the level found in part (a), how much energy must be available in a single collision?

(c) If such energetic collisions are to be affected simply by heating the gas until the average kinetic energy equals the desired upward energy jump, what temperature would be required? (This explains why heating is an impractical way to observe the hydrogen spectrum. Instead, the atoms are ionized by strong electric fields, as is the air when a static electric spark passes through.)

Energy levels (also called electron shells) are fixed distances from the nucleus of an atom where electrons can reside.

The energy required for the transition from one state to another state can be calculated by calculating the difference between the energies of those states.

If you have four downward lines that come to an end at n = 2 , there had better be electrons raised to the n = 3 , n = 4 , n = 5 , and n = 6 levels. Where is the principal quantum number.

Hence, the electrons should be exited up to at least n = 6 level.

If electrons jump very quickly emitting photons directly to the ground state then they must be raised from ground state to n = 6.

As you know that energy of a state is,

${\mathrm{E}}_{\mathrm{n}}=-\frac{13.6\mathrm{eV}}{{\mathrm{n}}^2}\left(\mathrm{n}=1,2,3,............\right)$

Where, n is the principal quantum number.

Hence, energy required for $1\to 6$transition is given by,

$\begin{array}{l}{\mathrm{E}}_{1\to 6}={\mathrm{E}}_6-{\mathrm{E}}_1\\ =(-13.6\mathrm{eV})\left(\frac{1}{6^2}-\frac{1}{1^2}\right)\\ =13.2\mathrm{eV}\\ \end{array}$

Hence, $13.2\mathrm{eV}$energy should be available in a single collision.

As you know that, Temperature required can be calculated with the help of the following equation.

Change in Energy,

$∆\mathrm{E}=\frac{3}{2}\mathrm{kBT}$

Where, kB is the Boltzmann constant and T is the temperature.

$$\begin{gathered} 13.2\times 1.6\times {10}^{-19}\mathrm{J}=\frac{3}{2}\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\mathrm{T} \\ \mathrm{T}=\frac{2\times 13.2\times 1.6\times {10}^{-19}}{3\times 1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}} \\ \mathrm{T}\cong {10}^5\mathrm{K} \end{gathered}$$

Hence, ${10}^5\mathrm{K}$ Temperature will be required for the collision.