Question

A Hydrogen atom is 5.3 x 10-11 m in radius. Use uncertainty Principle to estimate the minimum energy an electron can have in this atom.

Answer 1

minimum energy of an electron can have in this atom is 3.4 eV

It has given that a hydrogen atom is 5.3 × 10¯¹¹ m in radius.

we have to find the minimum energy of an electron in this atom using Uncertainty Principle

from Heisenberg's uncertainty principle,

∆P × ∆x ≥ h/4π

here, take ∆x = 5.3 × 10¯¹¹ m , h = 6.63 × 10¯³⁴ Js

now ∆P ≥ h/4π∆x

= (6.63 × 10¯³⁴ Js)/(4 × 3.14 × 5.3 × 10¯¹¹ m)

= 9.9 × 10^-25 Kgm/s .......(1)

you obviously know, an electron whose linear momentum is this order [ eq (1) ] behaves like a classical particle. so the kinetic energy of electron is K.E = P²/2m

here, P ≥ 9.9 × 10^-25 Kgm/s , m = 9.1 × 10¯³¹ kg

so, K.E = p²/2m ≥ (9.9 × 10^-25)²/(2 × 9.1 × 10¯³¹) = 5.4 × 10^-19 J

we know, 1eV = 1.6 × 10^-19 J

so, K.E ≥ (5.4 × 10^-19/1.6 × 10^-19) = 3.4 eV

so the kinetic energy in the lowest energy level of hydrogen atom is actually 3.4 × 2² = 13.6 eV