Hydrogen atom is bombarded with electrons accelerated through a potential difference of V, which causes excitation of hydrogen atoms. If the experiment is being performed at T=0 K, the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be α/10 V, where α= __________.

Solution

Physics sir · 3 steps

A Balmer line needs an electron falling to n = 2, so first push it up to at least n = 3.

1. What Balmer needs. Balmer lines come from jumps that end on n = 2. At T = 0 K every atom starts in the ground state n = 1, so the bombarding electron must lift it to at least n = 3. Only then can it fall from n = 3 to n = 2.

   Δ E = 13.6(1/1²-1/3²) eV = 13.6 × 8/9 eV = 12.09 eV

2. Energy to voltage. An electron accelerated through V volts gains eV of kinetic energy. To supply 12.09 eV it needs V = 12.09 V.
3. So. α/10 = 12.09, so α = 120.9, which rounds to 121.

The common slip: 102. Exciting the atom only to n = 2 gives 10.2 V and α = 102. But an electron at n = 2 can only drop to n = 1, which is a Lyman line.

The idea to keep. To see a line of any series, the atom must first reach a level above that series' lower level. For Balmer, starting from n = 1, the minimum excitation is to n = 3.

Answer from the NTA final answer key. Solution by PrepWiser.