A Hydrogen like atom of atomic Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition of quantum state n, a photon energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) of this atom. Also calculate the maximum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is - 13.6 eV. Solution :- Let the ground state in eV be E1 then E2n - E1 = 204 eV => - E1 = 204 eV and E2n- En = 40.8 ……………….. (i) = 40.8 eV, ;E1 = 40.8 eV …………………………….. (ii) from (ii) E1 = - n2(40.8) eV E1 = - (2)2(40.8) eV => E1 = - 217.6 eV E1 = - 13.6 Z2, z2 = => Z = 4 Emin = E2n - E2n - 1 = [here n = 2] Emin = 10.58 eV
A moving hydrogen atom makes a head on collision with a stationary hydrogen atom. Before collision both atoms are in ground state and after collision they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excitation state. Solution:- Let K be the kinetic energy of the moving hydrogen atom and K’, the kinetic energy of combined mass after collision.
from conservation of linear momentum, P = P’ or or, K = 2K’……………………………………………………………………………. (i) from conservation of energy, K = K’ + E …………………………………..(ii) Solving Eqn. (i) and (ii), we get E = K/2 Now, minimum value of E for hydrogen atom is 10.2 eV or The minimum kinetic energy of moving hydrogen is 20.4 eV
- E1 = 204 eV and E2n- En = 40.8 ……………….. (i)
= 40.8 eV, ;E1
= 40.8 eV …………………………….. (ii)
n2(40.8) eV
E1 = - 13.6 Z2,
=> Z = 4
E …………………………………..(ii)
