Radius of first orbit of a Bohr's atom is `0.265Å` When this atom makes a transition from first excited state to ground state, it emits a wavelength of `300 Å` . Find the ionisation energy for this atom in Rydberg's. (Take, hc `= 12375 eVÅ` and radius of hydrogen atom in ground state, `r_H=0.53Å`)

Answer

4

Explanation

[Ionisation energy=4 Ryd]
Note 1 Rydberg `=2.2 xx 10^(-18)` J = Rhc.
Radius of an atom `=r_H/Z = (0.53Å)/Z`
Now, `= (0.53xx10^(-10))/Z = 0.265 xx10^(-10), " " implies Z = (0.53)/(0.265) = 2`
Now, transition energy `DeltaE=E_2-E_1 = E_1/2^2-E_1`
`implies -3/4 E_1 = (hc)/(lamda)=(12375)/300, implies E_(1) = (-4 xx12375)/(3xx300)=-55eV`
So, ionisation energy `=55 eV = (55xx1.6 xx10^(-19))/(2.2 xx10^(-18))` Ryd = 4 Ryd