Question

Assume that the electron in He + ion is excited to the second orbit (n = 2). Calculate the
following:
(i) the radius of the orbit
(ii) the velocity of the electron
(iii) the potential energy of the electron, and
(iv)
the kinetic energy of the electron.
Hint: Use equations derived for hydrogen atom in Unit 1.

Answer 1

Answer on Question #54368 – Chemistry – General chemistry

Assume that the electron in He + ion is excited to the second orbit (n = 2). Calculate the following:

(i) the radius of the orbit

(ii) the velocity of the electron

(iii) the potential energy of the electron, and

(iv) the kinetic energy of the electron.

Hint: Use equations derived for hydrogen atom in Unit 1.

Solution

(i) the radius of the orbit

$$r = \frac {h ^ {2} n ^ {2}}{4 \pi m k Z e ^ {2}} = \frac {(6 . 6 \cdot 1 0 ^ {- 3 4}) ^ {2} \cdot (2) ^ {2}}{4 \pi \cdot 9 . 1 \cdot 1 0 ^ {- 3 1} \cdot 9 \cdot 1 0 ^ {9} \cdot 2 (1 . 6 \cdot 1 0 ^ {- 1 9}) ^ {2}} = 3. 3 \cdot 1 0 ^ {- 1 0} m.$$

(ii) the velocity of the electron

$$v = \frac {2 \pi k Z e ^ {2}}{h n} = \frac {2 \pi \cdot 9 \cdot 1 0 ^ {9} \cdot 2 (1 . 6 \cdot 1 0 ^ {- 1 9}) ^ {2}}{6 . 6 \cdot 1 0 ^ {- 3 4} \cdot 2} = 2. 2 \cdot 1 0 ^ {6} \frac {m}{s}.$$

(iii) the potential energy of the electron

$$P E = - \frac {k Z e ^ {2}}{r} = - \frac {9 \cdot 1 0 ^ {9} \cdot 2 (1 . 6 \cdot 1 0 ^ {- 1 9}) ^ {2}}{3 . 3 \cdot 1 0 ^ {- 1 0}} = - 1. 4 \cdot 1 0 ^ {- 1 8} J.$$

(iv) the kinetic energy of the electron

$$K E = \frac {1}{2} \frac {k Z e ^ {2}}{r} = \frac {1}{2} \frac {9 \cdot 1 0 ^ {9} \cdot 2 (1 . 6 \cdot 1 0 ^ {- 1 9}) ^ {2}}{3 . 3 \cdot 1 0 ^ {- 1 0}} = 7 \cdot 1 0 ^ {- 1 9} J$$

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