Question

The ionization energy of a hydrogen like bohr atom is `4` Rydbergs (i) What is the wavelength of the radiation emitted when the electron jumps from the first excited state to the ground state ?(ii) what is the radius of the orbit for this atom ?

Answer 1

The energy of the electron in hydrogen-like atom in nth orbit is
`E_(n) = (Z^(2) Rhc)/(n^(2))`
We have `Rhc = 1` rydberg.
The ionization energy
`E_(oo) - E_(1) = Z^(2) Rhc = 4 rydberg`
`:. Z^(2) = (4 rydberg)/(R h c) = (4 rydberg)/(1 rydberg) = 4`
`:. Z = 2`
a. The energy required to excite the electron from `n = 1` to `n = 2` is given by
`E_(2) - E_(1) = - (Z^(2) R h c)/(2^(2)) - ((-Z^(2) R h c)/(1^(2)))`
`= Z^(2) R h c (1 - (1)/(4))`
`= (3)/(4) Z^(2) R h c = (3)/(4) xx 4 Rydberg`
`= 3Rydberg`
If `lambda` is the wavelength of radiation emitted , then
`(h c)/(lambda) = 3 Rydberg`, i.e. `lambda = (h c)/((3 Rydberg))`
`:. lambda = (6.63 xx 10^(-34) xx 3 xx 10^(8))/(3 xx 2.2 xx 10^(-18)`
`= 301.4 xx 10^(-10) m = 301.4 Å`
b. Radius of first Bohr orbit `r_(1) = ((epsilon_(0) h^(2)// pi me^(2)))/(Z)`
`= (Radius of first bohr orbit of hydrogen)/(Z)`
`= (5 xx 10^(-11))/(2)`
2.5 xx 10^(-11) m`