Chapter 28, Problem 25P

(a)

To determine

The principal quantum number.

Answer to Problem 25P

The answer is $\text{6}$

Explanation of Solution

Given info: Hydrogen atom is in the state: $\text{6s}$

Formula used:

The principal quantum number $n$ can take on any integer value $\left(1,2,3,\mathrm{...}\right)$ and also, corresponds to the quantum number of the old Bohr theory.

Calculation:

According to the condition of principal quantum number, the principal quantum number is $\text{6}$

(b)

To determine

The energy of the state.

Answer to Problem 25P

The answer is $\text{-0}\text{.378}\text{eV}$

Explanation of Solution

Given info: Hydrogen atom is in the state: $\text{6s}$

The principal quantum number is $\text{6}$

Formula used:

Energy levels of $\text{H}$ atom are given by ${\text{E}}_{\text{n}}\text{=-}\frac{\text{13}\text{.6}}{{\text{n}}^{\text{2}}}\text{eV}$

Where, $\text{n}$ is the principal quantum number

Calculation:

The energy of the state is ${\text{E}}_{\text{n}}\text{=-}\frac{\text{13}\text{.6}}{{\text{n}}^{\text{2}}}\text{eV}\mathrm{...}\left(\text{1}\right)$

Substitute the values in equation $\left(\text{1}\right)$

  ${\text{E}}_{\text{6}}\text{=}\text{-}\frac{\text{13}\text{.6}\text{eV}}{{\text{6}}^{\text{2}}}\text{= -0}\text{.378}\text{eV}$

(c)

To determine

The orbital angular momentum and its quantum number.

Answer to Problem 25P

The orbital angular momentum and its quantum number is $0$ .

Explanation of Solution

Given info: Hydrogen atom is in the state: $\text{6s}$

The principal quantum number is $\text{6}$

Formula used:

The orbital quantum number $l$ can take on values from $0$ to $\text{n-1}$ .

The actual magnitude of the angular momentum $\text{L}$ is related to the quantum $l$ by

  $\text{L=}\sqrt{l\left(l+1\right)}h$ or $\text{L = h}{\left[l\left(l+1\right)\right]}^{\frac{1}{2}}$

Where, $l$ is the orbital quantum number and $h$ is Planck’s constant, $1.055\times {10}^{-34}J\cdot s$

Calculation:

According to the condition of quantum number, the $\text{"s"}$ sub shell has $l=0$ .

The magnitude of the angular momentum depends on $l$ only.

Substitute the given and values in expression $\text{L = h}{\left[l\left(l+1\right)\right]}^{\frac{1}{2}}$

  $\text{L = }\left(\text{1}{\text{.055×10}}^{\text{-34}}\text{J}\cdot \text{s}\right){\left[\text{0}\left(\text{0+1}\right)\right]}^{\frac{\text{1}}{\text{2}}}=0$

(d)

To determine

The possible values for the magnetic quantum number.

Answer to Problem 25P

The answer is $0$ .

Explanation of Solution

Given info: Hydrogen atom is in the state: $\text{6s}$

The principal quantum number is $\text{6}$

Formula used:

The orbital quantum number $l$ can take on values from $0$ to $\text{n-1}$ .

The magnetic quantum number, $m_l$ , can take on the integer values from $-l$ to $+l$

Calculation:

For each value of $l$ the value of $m_l$ can range from $-l$ to $+l$ . So, $m_l=0$ .