Question

With increasing quantum number the energy difference between adjacent level in atoms

Answer 1

With increasing quantum number the energy difference between adjacent levels in atoms.

Solution.

The energy levels of an electron around a nucleus:

$$E _ {n} = - \frac {m e ^ {4} Z ^ {2}}{8 \varepsilon_ {0} ^ {2} h ^ {2} n ^ {2}};$$

- $m$ - the rest mass of the electron;

- $e$ - the elementary charge;

- $Z$ - the atomic number;

- $\varepsilon_0$ - the permittivity of free space;

- $h$ - the Planck constant;

- $n$ - the principal quantum number.

The energy difference between adjacent levels with quantum numbers $n + 1$ and $n$:

$$\begin{array}{l} \Delta E _ {n + 1, n} = E _ {n + 1} - E _ {n} = - \frac {m e ^ {4} Z ^ {2}}{8 \varepsilon_ {0} ^ {2} h ^ {2}} \left(\frac {1}{(n + 1) ^ {2}} - \frac {1}{n ^ {2}}\right) = \\ = \frac {m e ^ {4} Z ^ {2}}{8 \varepsilon_ {0} ^ {2} h ^ {2}} \left(\frac {1}{n ^ {2}} - \frac {1}{(n + 1) ^ {2}}\right) = \frac {m e ^ {4} Z ^ {2}}{8 \varepsilon_ {0} ^ {2} h ^ {2}} \left(\frac {2 n + 1}{n ^ {4} + 2 n ^ {3} + n ^ {2}}\right). \end{array}$$

The energy difference between adjacent levels with quantum numbers $n + 2$ and $n + 1$:

$$\begin{array}{l} \Delta E _ {n + 2, n + 1} = E _ {n + 2} - E _ {n + 1} = - \frac {m e ^ {4} Z ^ {2}}{8 \varepsilon_ {0} ^ {2} h ^ {2}} \left(\frac {1}{(n + 2) ^ {2}} - \frac {1}{(n + 1) ^ {2}}\right) = \\ = \frac {m e ^ {4} Z ^ {2}}{8 \varepsilon_ {0} ^ {2} h ^ {2}} \left(\frac {1}{(n + 1) ^ {2}} - \frac {1}{(n + 2) ^ {2}}\right) = \frac {m e ^ {4} Z ^ {2}}{8 \varepsilon_ {0} ^ {2} h ^ {2}} \left(\frac {2 n + 3}{n ^ {4} + 6 n ^ {3} + 13 n ^ {2} + 12 n + 4}\right). \end{array}$$ $$\frac {2 n + 3}{n ^ {4} + 6 n ^ {3} + 13 n ^ {2} + 12 n + 4} < \frac {2 n + 1}{n ^ {4} + 2 n ^ {3} + n ^ {2}}, \text{ then } \Delta E _ {n + 2, n + 1} < \Delta E _ {n + 1, n}.$$

With increasing quantum number the energy difference between adjacent levels in atoms decreases.

Answer: With increasing quantum number the energy difference between adjacent levels in atoms decreases.