Chapter 39: Q4P (page 1215)

An electron, trapped in a one-dimensional infinite potential well 250 pm wide, is in its ground state. How much energy must it absorb if it is to jump up to the state with $n = 4$ ?

Short Answer

Expert verified

90.3 eV

Step by step solution

Given data

The width of the potential well is

$L = 250 pm = 0.250 nm n = 4$

Energy in a potential well

The $n^{t h}$ state energy of a particle of mass in an infinite potential well of width L is

$E_{n} = (\frac{n^{2} h^{2}}{8 m L}) . . . . . . . . . . (1)$

Determining the energy required to jump from ground state to the fourth state

Here h is the Planck's constant having value

$h = 6.626 \times 10^{- 34} J . s and c = 2.998 \times 10^{8} m / s h_{c} = \frac{(6.626 \times 10^{- 34} J . s) (2.998 \times 10^{8} m / s)}{(1.602 \times 10^{- 19} J / eV) (10^{- 9} m / nm)} = 1240 eV . nm$

Using the value for an electron from Table 37-3 $(511 \times 10^{3} eV)$ , Eq.39 – 4 can be rewritten as

$E_{n} = \frac{n^{2} h^{2}}{8 m L^{2}} = \frac{n^{2} {(h_{c})}^{2}}{8 (m c^{2}) L^{2}}$

The absorbed energy is

$∆ E = E_{4} - E_{1} = \frac{(4^{2} - 1^{2}) {(h_{c})}^{2}}{8 (m_{e} c^{2}) L^{2}} = \frac{15 {(1240 eV . nm)}^{2}}{8 (511 \times 10^{3} eV) {(0.250 nm)}^{2}} = 90.3 eV$