Hydrogen gas can be placed inside a strong magnetic field B=12T. The energy of 1s electron in hydrogen atom is 13.6 eV ( 1eV= 1.6*10» J ). a) What is a wavelength of radiation corresponding to a transition between 2p and 1s levels when magnetic field is zero? b) What is a magnetic moment of the atom with its electron initially in s state and in p state? c) What is the wavelength change for the transition from p- to s- if magnetic field is turned on?

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Hydrogen gas can be placed inside a strong magnetic field B=12T. The energy of 1s
electron in hydrogen atom is 13.6 eV ( 1eV= 1.6*10 J ).
a) What is a wavelength of radiation corresponding to a transition between 2p and 1s levels when
magnetic field is zero?
b) What is a magnetic moment of the atom with its electron initially in s state and in p state?
c) What is the wavelength change for the transition from p- to s- if magnetic field is turned on?
Transcribed Image Text:Hydrogen gas can be placed inside a strong magnetic field B=12T. The energy of 1s electron in hydrogen atom is 13.6 eV ( 1eV= 1.6*10 J ). a) What is a wavelength of radiation corresponding to a transition between 2p and 1s levels when magnetic field is zero? b) What is a magnetic moment of the atom with its electron initially in s state and in p state? c) What is the wavelength change for the transition from p- to s- if magnetic field is turned on?
Expert Solution
Step 1

The Bohr energy (En) of the nth-state of an H-atom can be expressed in terms of its ground-state (1s) energy (E0) as follows:

 

En=E0n2

 

Here, n is the principal quantum number. The ground-state binding energy is 13.6 eV whereas the actual total ground-state energy is – 13.6 eV.

Step 2

a)

Let h and c denote Planck’s constant and the light speed constant, respectively.

The H-atom’s 2s and 2p levels are degenerate. Hence, the above relation along with the Planck energy-wavelength relation can be used to determine the required wavelength (say λ0) as follows:

 

E2p-E1s=hcλ0λ0=hcE022-E012=44.14×10-15 eV·s3×108 m/s-3-13.6 eV=1.22×10-7 m or 1216.37 Ao               1 Ao=10-10 m

Step 3

b)

The magnetic moment (μ) of a state with the azimuthal quantum number (l) can be expressed by using the Bohr magneton (μB) as follows:

 

μ=-e2meLL=ll+1L^μ=-μBll+1L^                μB=e2me

 

Here, vector L denotes the orbital angular momentum vector, and e, me, and ħ are the electron’s charge, its rest mass, and the reduced Planck’s constant, respectively.

The negative sign of μ just implies that it is directed oppositely to L.

Step 4

Hence, evaluate the magnitudes of μ0 and μ1 corresponding to the s-state (l = 0) and the p-state (l = 1), respectively as follows:

 

μ0=μB00+1=0μ1=5.79×10-5 eV/T11+1=8.186×10-5 eV/T

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