In an hydrogen atom- the electron revolves around the nucleus in an orbit of radius nbsp- 0-53 -10- - 10m- Then the electrical potential produced by the nucleus at the position of the electron is

V = ( Kqr) = ( 9 ×10^9× 1.6 ×10^ 190.53 ×10^ 10) #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; = 27.16 ...

You visited us 23 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Electric Potential at a Point in Space Due to a Point Charge Q

In an hydroge...

Question

In an hydrogen atom, the electron revolves around the nucleus in an orbit of radius $0.53 \times 10^{- 10} m .$ Then the electrical potential produced by the nucleus at the position of the electron is

Open in App

Solution

$V = (\frac{K q}{r}) = (\frac{9 \times 10^{9} \times 1.6 \times 10^{- 19}}{0.53 \times 10^{- 10}})$
$= 27.16 V$

Suggest Corrections

Similar questions

An electron revolves around the nucleus of hydrogen atom in a circle of radius $5 \times 10^{- 11} m$ . The intensity of electric field at a point in the orbit of the electron is:

Q. In hydrogen atom the electron is making

6.6 \times 10^{15}

rev/s around the nucleus of radius 0.53 A. The magnetic field produced at the centre of the orbit is nearly

An electron is circulating around the nucleus of the hydrogen atom in a circular orbit of radius 5.3×10^-11 m calculate the electric potential at this radius and electric potential energy of the atom in EV.

Q. An electron in an atom revolves around the nucleus in an orbit of radius

0.53 \circ A

. Calculate the equivalent magnetic moment, if the frequency of revolution of electron is

6.8 \times 10^{9}

MHz(Given :

e = 1.6 \times 10^{- 19}

)

Q. An electron revolves in an orbit of hydrogen atom with the speed of

10^{5} m / s

. The magnetic field produced at the nucleus due to the motion of electron is (Given, radius of orbit

= 0.2 o A, μ_{0} = 4 π \times 10^{- 7} H / m

Join BYJU'S Learning Program

In an hydrogen atom, the electron revolves around the nucleus in an orbit of radius 0.53×10−10m. Then the electrical potential produced by the nucleus at the position of the electron is

V=(Kqr)=(9×109×1.6×10−190.53×10−10) =27.16V

In an hydrogen atom, the electron revolves around the nucleus in an orbit of radius $0.53 \times 10^{- 10} m .$ Then the electrical potential produced by the nucleus at the position of the electron is

$V = (\frac{K q}{r}) = (\frac{9 \times 10^{9} \times 1.6 \times 10^{- 19}}{0.53 \times 10^{- 10}})$
$= 27.16 V$