IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

raman_1290

Nice question yaar

the energy of electron in first shell of H-atom = 13.6 eV

or (1/2)mv² = 2.178 X 10^-18 J

or v² = (2 X 2.178 X 10^-18 )/(9.1 X 10^-31)

v² = 4.787 X 10¹²

v = 2.188 X 10⁶m/s

Now if it is true that the electrostatic force of attraction between the nucleus and electron provides the centripetal force to the electron for revolving around the nucleus, then

ke²/r² = mv²/r

r = Bohr?s radius = 0.529 X 10^-10 m

or ke²/r = mv²

v² = ke²/mr

= (9 X 10⁹)(1.6 X 10^-19)²/ (9.1 X 10^-31)(5.29 X 10^-11)

= 4.786 X 10¹²

v = 2.187 X 10⁶ m/s

Thus, , velocity of electron in first orbit of H atom comes out to be the same from both the experiments. Hence proved that the electrostatic force of attraction between the nucleus and electron provides the centripetal force to the electron for revolving around the nucleus.

Cheers!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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