Question

What is the ionization energy of Li+2 i.e., the energy for the process Li+2 --> Li+3 + e- , ans: 1.96 x 10^-17 J.
and Upon absorption of a photon the deBroglie wavelength of a sodium atom is 10^-4 m. What is the speed of the atom? ans: 1.74 x 10^-4 Please explain.

Answer 1

1) We know that 1/λ = Rz² [ 1/n1^2- 1/n2^2]

λ = wavelength

R = Rydberg constant = 1.097 x 10⁷ m^-1

   Z = atomic number

Atomic number of Li, Z = 3

Ionization means removing of electron from n= 1 to n= ∞

Therefore, n1= 1, n2= ∞

1/λ = Rz² [ 1/n1^2- 1/n2^2]

E = hc/λ = hcRz² [ 1/n1^2- 1/n2^2]

= (6.626 x 10^-34 J.s) (3 x 10⁸ m/s) (1.097 x 10⁷ m^-1) (3²)[ 1/1²- 0]

= 1.96 x 10^-17 J

Therefore, Ionization energy =  1.96 x 10^-17 J

2)    deBroglie wavelength λ = h/mv

v = h/λm

= 6.626 x 10^-34 J.s / 10^-4 m x 9.1 x 10 ^-31 kg

= 7.28 m/s

speed of atom =  7.28 m/s