Name: The Classical Model of an Atom
Uploaded: 2019-07-19
Duration: 1 min 51 s
Description: The current model of the atom involves a dense nucleus surrounded by a probabilistic "cloud" of electrons. Do you know the classical model of an atom? Watch...

Question

Find the radius of the hydrogen atom in its ground state. Also calculate the velocity of the electron in

n = 1

orbit. Given

h = 6.62 \times 10^{- 34} Js, m = 9.1 \times 10^{- 31} kg, e = 1.6 \times 10^{- 19} C, k = 9 \times 10^{9} {Nm}^{2} C^{- 2}

.

Accepted Answer

We know that the radius of a stationary orbit of a hydrogen atom is given by:

$r = \frac{n^{2} h^{2}}{4 π^{2} m K e^{2}}$

Given: $h = 6.62 \times 10^{- 34} Js, m = 9.1 \times 10^{- 31} kg, e = 1.6 \times 10^{- 19} C, k = 9 \times 10^{9} {Nm}^{2} C^{- 2}, n = 1$

Putting these values in the formula, we get,

$r = \frac{{(1)}^{2} {(6.6 \times 10^{- 34})}^{2}}{4 \times {(3.14)}^{2} \times 9.1 \times 10^{- 31} \times 9 \times 10^{9} \times {(1.6 \times 10^{- 19})}^{2}} r = 5.3 \times 10^{- 11} m$

The formula for calculating velocity of electron in a stationary orbit of a hydrogen atom:

$v = \frac{2 π K e^{2}}{n h}$

Given: $h = 6.62 \times 10^{- 34} Js, n = 1, e = 1.6 \times 10^{- 19} C, k = 9 \times 10^{9} {Nm}^{2} C^{- 2}$

Putting these values in the formula,

$v = \frac{2 \times 3.14 \times 9 \times 10^{9} \times {(1.6 \times 10^{- 19})}^{2}}{1 \times 6.6 \times 10^{- 34}} v = 2.2 \times 10^{6} m / s$

MODEL QUESTIONS (IPE PATTERN)

Telangana Board Physics Solutions for Exercise - MODEL QUESTIONS (IPE PATTERN)

Questions from MODEL QUESTIONS (IPE PATTERN) with Hints & Solutions

Learn and Prepare for any exam you want !