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A doubly ionized lithim atom is hydrogen...

A doubly ionized lithim atom is hydrogen like with atomic number 3. Find the wavelength of the radiation required to excite the electron in ` Li^(2+)` from the first to the third Bohr orbit. The ionization energy of the hydrogen atom is 13.6 eV. (Take hc=1240 eV nm)

A

11.4 nm

B

10.7 nm

C

12.7 nm

D

13.4 nm

Text Solution

Verified by Experts

The correct Answer is:
C
The energy of `n^("th")` orbit of a hydrogen -like atom is given by `E_(n) = (13.6 Z^(2))/(n^(2)) eV`
For ` Li^(2+) , Z =3 , E_(n) = (122.4)/(n^(2)) eV`
` E_(1) = (122.4)/(I)^(2) = -122.4 e V and E_(3) = - (122.4)/((3)^(2)) =-13.6 eV`
` therefore Delta E = E_(3)-E_(1) = 108.8 eV`
THe corresponding wavelength is
` lamda = (hc)/(Delta E) = (1240 eV nM)/( 108.8 eV) = 11.4 nm`
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