Question

Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH=1.06×10−10m d H = 1.06 × 10 − 10 m , into the answer box.

Answer 1

Answer:

The diameter of the hydrogen $\mathbf{d =1.0605 \times 10^{-10}\ m}$

Explanation:

From the given information:

Using the concept of Bohr's Model, the equation for the angular momentum can be expressed as:

$L = \dfrac{nh}{2 \pi}$

Where the generic expression for angular momentum is:

L = mvr.

replacing the value of L into the previous equation, we have:

$mvr= \dfrac{nh}{2 \pi}$

$v= \dfrac{nh}{2 \pi mr}$ ----- (1)

The electron in the hydrogen atom posses an electrostatic force which gives a centripetal force.

$\dfrac{ke^2}{r^2} = \dfrac{mv^2}{r}$   ----- (2)

replacing the value of v in equation (1) into (2), and taking r as the subject of the formula, we have:

$\dfrac{ke^2}{r} = m (\dfrac{nh}{2 \pi mr})^2$

$ke^2=\dfrac{n^2h^2}{4 \pi^2 mr}$

$r =\dfrac{n^2h^2}{4 \pi^2 mke^2}$

For ground-state n = 1

$h = (6.625 \times 10^{-34} \ J.s)^2$

$m =( 9.1 \times 10^{-31} \ kg)(9 \times 10^9 \ N .m^2/C^2)$

$Ke = (1.6 \times 10^{-19} \ C)^2$

$r =\dfrac{(1)^2(6.625 \times 10^{-34})^2}{4 \pi^2 (9.1 \times 10^{-31} )(9 \times 10^9 ) (1.6 \times 10^{-19})^2}$

$r =\dfrac{4.3890625 \times 10^{-67}}{8.27720295 \times 10^{-57}}$

$\mathbf{r = 5.3025 \times 10^{-11} \ m}$

Therefore, the diameter of hydrogen d = 2r

$\mathbf{d = ( 2 \times 5.3025 \times 10^{-11} \ m})}$

$\mathbf{d =1.0605 \times 10^{-10}\ m}}$