Question

Consider the first three energy levels of hydrogen (n = 1, 2, 3).

a) What photon energies can be observed from transitions between these levels? Label these in increasing order as E₁, E₂, and E₃.

b) A hydrogen atom which is initially in the n = 2 level collides with an aluminum atom in its ground state (the kinetic energy of the collision is nearly zero). The hydrogen can drop to the n = 1 level and ionize the aluminum atom. If the ionization energy of aluminum is 6 eV, what energy will the emitted electron have (in eV)?

c) Suppose you have mystery atom, and collide it with hydrogen atoms prepared in the n = 2 level, observing electrons with energy 6.3 eV. What is the ionization energy of the mystery atom?

Answer 1

b. E_{photon emitted from H atom,n=2 to n=1)}= E₂-E₁ = 10.2 eV . Now this photon will ionize the aluminum atom.

Using the relation,

K.E = E_photon - Ionization energy

K.E. = 10.2 eV - 6 eV = 4.2eV

c.

According to what you have asked,the mystery atom is getting ionized. So The H atom in n=2 will fall to n=1 after ionizing the mystery atom. Here,again

E_{photon emitted from H atom,n=2 to n=1)}= E₂-E₁ = 10.2 eV

Using the relation,

K.E = E_photon - Ionization energy

6.3 eV = 10.2 eV - Ionization energy

Hence, Ionization Energy= 10.2eV -6.3 eV = 3.9 eV