Question

Consider the ground state energy of the hydrogen atom E₀.

Enter the expression you use to verify that the ground state energy is 13.6 eV? Use fundamental constants e, m_e, k and h.

E₀ = ?

Answer 1

Ground state energy of hydrogen atom is given by

E₀ = ( 2 ?² e⁴ m_e k² ) / h²

= { 2 × 3.14² × ( 1.6 × 10^-19 )⁴ ×   9.1 × 10^-31 ×( 9 × 10⁹ )² } / ( 6.626 × 10^-34 )²

= ( 95256.72 × 10^-21 / 43.90 )

= 2169.85 × 10^-21 J

We know that 1 eV = 1.6 × 10^-19 J , We can write

E₀ = ( 2169.85 × 10^-21 ) / ( 1.6 × 10^-19 )

= 13.56 eV

= - 13.6 eV ( approximately )

Negative sign is used to indicate that electron is bound to the nucleus.

We can also use the following formula for finding ground state energy of the electron.

E₀ = m* e⁴ / ( 8 * ₀² * n² * h²)

where orbit n =1,

.   ₀ = 8.85 * 10^-12 C²/N.m²

h = 6.626 * 10^-34 J-s,

m = 9.1 * 10^-31 kg , e = 1.6 * 10^-19 C

on substituting we get,

E₀ = (1.6 * 10^-19)⁴ * (9.1 * 10^-31)/8* 1* (8.85 * 10^-12)² * (6.626 * 10^-34)²

= 21.76 * 10^-19 J

= 21.76 * 10^-19/(1.6 * 10^-19) eV

= 13.6 eV

So ground state energy of hydrogen atom = 13.6 eV

E_n = -13.6/n² eV where n= 1,2,3..................