The radii of two of the first four Bohr orbits of the hydrogen atom are in the ratio 1:4. The energy difference between them may be

Options

A. Either 12.09 eV or 3.4 eV
B. Either 2.55 eV or 10.2 eV
C. Either 13.6 eV or 3.4 eV
D. Either 3.4 eV or 0.85 eV

Answer

B

Explanation

Using Bohr radius formula for `n^(th)` orbit:
`r_n = n^2 (h^2/(4pi^2 kme^2)). 1/Z`
`(h^2/(4pi^2 kme^2)) = a_0 = 0.529 Å`,
This `a_0` is equal to the first Bohr radius for the hydrogen atom.
First four orbits radius ratio for hydrogen atom: `r_1 : r_2 : r_3 : r_4`
`1 : 4:9:16`
In the first four orbits, there are two possible sets having the radius ratio 1:4. These sets are either `(n_1 = 1, n_2 = 2) " or " (n_1 = 2, n_2 = 4)`.
The energy difference between two orbits is given by ni-lower and `n_2` higher orbit)
`DeltaE_(n_1 to n_2) = 13.6 xx (1/(n_1^2) -1/(n_2^2) ) eV`
The energy difference between first and second orbit:
`DeltaE_(1 to 2) = 13.6 (1/1^2 -1/2^2 ) eV = 13.6 xx 3/4= 10.2 eV`
The energy difference between the second and fourth orbit,
`DeltaE_(2 to 4) =13.6xx(1/2^2 -1/4^2)=13.6xx3/16 =2.55 eV`