Question

A hydrogen atom is in the n = 3 state. Determine, according to quantum mechanics, (a) the total energy (in eV) of the atom, (b) the magnitude of the maximum angular momentum the electron can have in this state, and (c) the maximum value that the z component Lz of the angular momentum can have.

Answer 1

For a hydrogen atom in the n = 3 state, the total energy is -1.51 eV, the magnitude of the maximum angular momentum is √6 ħ, and the maximum value of the z component of angular momentum is 2ħ.

Step-by-step explanation:

A hydrogen atom in the n = 3 state has several key quantum mechanical properties. According to quantum mechanics:

Total energy (E) of the atom in an energy level is given by the formula E = -13.6 eV / n². For n = 3, this gives E = -13.6 eV / (3²) = -1.51 eV.

The magnitude of the maximum angular momentum (L) that the electron can have is defined by L = √(l(l + 1))ħ, where the maximum value of l for n = 3 is 2. Plugging in the values, we get L = √(2(2 + 1))ħ ≈ √6 ħ.

The maximum value of the z component of the angular momentum (Lz) can be given by mħ, where m is the magnetic quantum number and can range from -l to l. For l = 2, the maximum m is 2, hence the maximum Lz = 2ħ.

The student must understand that these values arise from the quantum mechanical description of atoms, which differs from the earlier Bohr model primarily in its allowance for quantized angular momenta.

Answer 2

Step-by-step explanation:

Given that,

Number of state = 3

(a). We need to calculate the total energy

Using formula of the total energy

`E=-(13.6*z^2)/(n^2)`

Put the value of n

`E=(-13.6*1)/(3^2)`

`E=1.511\ eV`

(b). L= n-1= 3-1 = 2

Using formula of angular momentum

`J=(h)/(2\pi)√(l(l+1))`

`J=(6.6*10^(-34))/(2\pi)*√(2(2+1))`

`J=2.572*10^(-34)\ J-s`

(c). Maximum value l =2

We need to calculate the the angular momentum

`J=(2h)/(2\pi)`

`J=(2*6.6*10^(-34))/(2\pi)`

`J=2.100*10^(-34)\ J-s`

Hence, This is required solution.