Determine the gravitational attraction between the proton and the electron in a hydrogen atom,assuming that the electron describes a circular orbit with a radius of 0.53×10⁻¹⁰ m.mass of electron=9.1×10⁻³¹ kg,mass of proton=1.67×10⁻²⁷ kg

Physics class 11th problem solution  Physics first year problem solution  Gravitation problems solution

Transcript

00:00Question is, determine the gravitational attraction between the proton and the electron in an

00:15atom of hydrogen, assuming that electron describes a circular orbit of radius 0.53 into 10 raised

00:26power minus 10 meter. So, first of all, we will form a data, is the mass of proton and electrons

00:41are given, which are constant, is mass of electron is 9.11 into 10 raised power minus

00:5431 kilogram. And the mass of proton is 1.67 into 10 raised power minus 27 kilogram. And

01:13the distance between them is, which is given as 0.53 into 10 raised power minus 10 meter,

01:27is the value of capital G is given as 6.67 into 10 raised power minus 11 Newton into meter

01:39square per kilogram square. So, we have to find the gravitational attraction between proton

01:49and electron, means F. So, for solution, we will use the formula is, the gravitational attraction

02:04between proton and electron will be calculated by this formula, F is equal to G into mass of

02:14proton and electron into mass of proton divided by R square. Now, by putting these given values,

02:28the gravitational force will be obtained as, as the value of capital G is 6.67 into 10 raised

02:35power minus 11 into mass of electron is 9.11 into 10 raised power minus 31 into the mass of

02:44proton. So, the proton is 1.67 into 10 raised power minus 27 kilogram divided by R square, which is obtained

02:56is 0.53 into 10 raised power minus 10 whole square. So, the gravitational force of attraction will be obtained

03:11as, when we will multiply 6.67 into 10 raised power minus 10 raised power minus 10 raised power. So, the

03:12answer is, when we will multiply 6.67, 9.11 and 1.67, then we will get the answer is, 101.36 into, and as the

03:31bases are same in multiplication, then we will add the powers. So, these will be 10 raised power minus 11,

03:40minus 11, minus 31, minus 27 divided by the square of 0.53 will be 0.2809 into 10 raised power minus 20. The

04:05gravitational force will be 10 raised power minus 10 raised power minus 10 raised power minus 10 is 10 raised power

04:08minus 12. This gravitational force will be obtained as 101.36 into 10 raised power minus 11, minus 31, minus 27. It will

04:23to come is minus 69 divided by 0.2809 into 10th power minus 20. So, the gravitational

04:36force will be obtained as is 101.36 divided by 0.2809 then it will be obtained as 360.84

04:58into S in DVN when the bases are same then we will subtract the power of denominator from

05:09the power of numerator. So, minus 20 will become S plus 20 in numerator. So, the gravitational

05:20force will be obtained as 360.84 into 10th power minus 69 plus 20. It will be minus 49 Newton

05:38or we can also write it as when we will take the decimal up to 2 digits from right to left.

05:52So, the force will be obtained as 3.6084 into 10th power then we will add 2 with the power

06:07means when we will shift the decimal up to 2 digits from right to left then we will add

06:152 with the power. So, it will become as minus 47 Newton. So, this is our required answer.

06:29You can see you in order. You can leave the bottom of the grade and then it will be

06:38noon. So, everybody binnen the 10 so that you consider it. So, the power of the triangle and

06:41theness of the triangle ensemble is on the way our determination it will be it becomes

06:422 of 10 9 D or 4