In a hydrogen like atom the ionisation energy equals 4 times Rydberg's constant for hydrogen. What is the wavelength of radiations emitted when a jump takes place from the first excited state to the ground state? What is the radius of first Bohr's orbit?

Solution

The ionization energy E of a hydrogen-like Bohr atom of atomic number Z is given by,

E = - {Rz}^{2} = - \frac{2 {πk}^{2} z^{2} {me}^{4}}{n^{2} h^{2}}

where the Rydberg constant is

R = \frac{2 π^{2} {kme}^{4}}{n^{2} h^{2}} = 2.2 \times 10^{- 18} J

Given that,
Ionisation energy is equal to four times the rydberg constant.

i.e.,

E = 4 \times Rydberg constant = 4 R, we have

∴

4 R = {RZ}^{2}

or,

Z = 2

(i) Energy of radiation emitted (E) when the electron jumps from the first excited state to the ground state is given by,

E = {RZ}^{2} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = 4 R (1 - \frac{1}{4}) = 3 R = 3 \times 2.2 \times 10^{- 18} J

i.e.,

E = 6.6 \times 10^{- 18} J .

Wavelength of the radiation emitted,

λ = \frac{hc}{E} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{6.6 \times 10^{- 18}} = 3 \times 10^{- 8} m

(ii) Radius of the first Bohr orbit,

= \frac{Bohr radius of hydrogen atom}{Z} = \frac{5 \times 10^{- 11}}{2} = 2.5 \times 10^{- 11} m .

For Daily Free Study Material Join wiredfaculty