Question

Find the radius and energy of a He+ ion in the states (a) n = 1, (b) n = 4 and (c) n = 10.

Answer 1

Radius of He+ ion in states of

a. n=1 is 0.264 A°

b. n=4 is 4.232 A°

c. n=10 is 26.45 A°

Energy of He+ ion in states of

a. n=1 is -54.4 J

b. n=4 is -3.4 J

c. n=10 is -0.544 J

Formula :

• For calculating radius of ion is 0.529*n^2/Z and
• For calculation energy is -13.6 Z^2/n^2
• Where Z is the atomic number
• Here for He, Z is 2
• n = number of states

Answer 2

The radius and energy of a He+ ion a) In the state n = 1 is $r=0.265 A^{\circ} \text { and } e=-54.4 e V$, b) In the state n=4 is $r=4.24 A^{\circ} \text { and } e=-3.4 e V$, c) In the state n=10 is $r=26.5 A^{\circ} \text { and } e=-0.5 e V$.

Explanation:

The radius and Energy can be determined using the formula,

$r= \frac{\varepsilon_{0} h^{2} n^{2}}{Z e^{2} \pi m}$

$\Rightarrow$ $E_{n}=-\frac{13.6 Z^{2}}{n^{2}}$

By substituting n= 1 ,4 and 10 respectively the above values can be obtained easily.

a) In the state n=1,

$r=\frac{0.53 \times(1)^{2}}{2}$

$\Rightarrow E_{n}=\frac{-13.6 \times 4}{1}$

$=-54.4 \mathrm{ev}$

b) In the state n=4,

$r=\frac{0.53 \times 16}{2}=4.24 \hat{\mathrm{A}}$

$\Rightarrow E=\frac{-13.6 \times 4}{16}=-3.4 e V$

c) In the state n=10,

$r=\frac{0.53 \times 100}{2}$

$\Rightarrow \mathrm{E}=\frac{-13.6 \times 4}{100}=-0.544 e V$