Question

Consider a hydrogen atom in the n = 1 state. The atom is placed in a uniform B field of magnitude 2.5 T. Calculate the energy difference between the highest and lowest electronic energy levels in the presence of the B field. a. 9.3 X 10^-5 eV
b. 29 X 10^-5 eV
c. 11.6 X 10^-5 eV

Answer 1

Answer:

$E=29\times 10^{-5}eV$

Explanation:

For n-=1 state hydrogen energy level is split into three componets in the presence of external magnetic field. The energies are,

$E^{+}=E+\mu B$,

$E^{-}=E-\mu B$,

$E^{0}=E$

Here, E is the energy in the absence of electric field.

And

$E^{+} and E^{-}$ are the highest and the lowest energies.

The difference of these energies

$\Delta E=2\mu B$

$\mu=9.3\times 10^{-24}J/T$ is known as Bohr's magneton.

B=2.5 T,

Therefore,

$\Delta E=2(9.3\times 10^{-24}J/T)\times 2.5 T\\\Delta E=46.5\times 10^{-24}J$

Now,

$Delta E=46.5\times 10^{-24}J(\frac{1eV}{1.6\times 10^{-9}J } )\\Delta E=29.05\times 10^{-5}eV\\Delta E\simeq29\times 10^{-5}eV$

Therefore, the energy difference between highest and lowest energy levels in presence of magnetic field is $E=29\times 10^{-5}eV$