Question

If m is the mass of an electron, h is the Planck's
constant and a, is the radius of first orbit of
H-atom, then kinetic energy of electron in the
h2
third orbit of He* ion is given by
1
The value of x is

Answer 1

Answer:

32

the valu of x

Explanation:

the value of mass and atom is 32

Answer 2

Given : mass of electron is m, plank's constant is h , radius of first orbit of H - atom is a.

To find : The kinetic energy of electron in 3rd orbit of He⁺ ion is given by,

solution : according to Bohr's atomic theory,

radius of nth orbit is given by, R = a × n²/Z where a is radius of first orbit, Z is atomic no.

so radius of 3rd orbit of He⁺ ion is R = a × (3)²/2 = 4.5a

as we know, electrostatic force acting between electron and nucleus is balanced by centripetal force acting on the electron.

⇒K(Ze)(e)/R² = mv²/R

here Z = 2 [ for He] , R = 4.5a

⇒2Ke²/(4.5a) = mv²

⇒1/2 mv² = Ke²/4.5a

⇒kinetic energy = Ke²/4.5a

here K = 9 × 10^9 Nm²/C²

e = charge on electron = 1.6 × 10^-19

so, kinetic energy = (9 × 10^9 × (1.6 × 10^-19)²}/4.5a

= (5.12 × 10^-29)/a

kinetic energy of electron in 3rd orbit of He⁺ ion is given by (5.12 × 10^-29)/a