Question

An electron is in a box 2nm across.what is the lowest energy for the electron

Answer 1

Explanation:

i think it will be about 1 nm

Answer 2

When an elementary particle like electron moves inside one dimensional box of width L, the energy of the particle is given by E = $n^{2}h^{2} / 8mL^{2}$

Explanation:

• Eigen values of particle confined in one dimensional box is the values of energy levels of different states
• designated by quantum number n
• Energy level of nth state = En = [ $h^{2}$/(8m$L^{2}$ )]$n^{2}$
• where h = 6.626 × $10^{-34}$ J , planck's constant
• m = 9.109 × $10^{-31}$ kg , electron mass
• L = $10^{-10}$ m = dimension of box
• n is quantum state

• By substituting the values , we get, En = 6.025 × $10^{-18}$ $n^{2}$  J

• since 1 eV = 1.602 × $10^{-19}$ J , we express energy of states in eV as, En = (6.025 × $10^{-18}$ n2 ) / ( 1.602 × $10^{-19}$ ) eV
• En = 37.61 n2 eV
• Hence , E1 = 37.61 eV
• E2 = 37.61 × 4 = 150.44 eV
• E3 = 37.61 × 9 = 338.5 eV
• E4 = 37.61 × 16 = 601.76 eV