Question

Energy of electron trapped in 1-D infinite potential box of width 1 angstrom. Approximate energy of electron in 5th excited state is

60.2 eV

1355.2 eV

1844.6 eV

941.1 eV​

Answer 1

Explanation:

The probability that the electron is found in any interval is given by P=∫∣ψ∣

2

dx. where the integral is over the interval. If the interval width Δx is small, the probability can be approximated by P=∣ψ∣

2

Δx, where the wave functions is evaluated for the center of the interval say. For an electron trapped in an infinite well of width

L, the ground state probability density is

∣ψ∣

2

=

L

2

sin

2

(

L

πx

)

P=(

L

2Δx

)sin

2

(

L

πx

)

We take L=100 pm,x=90 pm, and Δx=5.0 pm. Then,

P=[

100 pm

2(5.0 pm)

]sin

2

[

100 pm

π(90 pm)

]=0.0095.

Answer 2

Answer:

The energy of an electron in the ground state in a one-dimensional infinite  potential well of width 2 Å is $1352eV$

Explanation:

• The nth state energy of a particle in an infinite potential well of width L is given by the formula

                                $E_{n} =\frac{n^{2} h^{2} }{8mL^{2} }$

       where,

       $n$-state of the particle

       $m$-the mass of the particle

       $h$-$6.26*10^{-34}Js$Planck's constant

      $L$-width of the potential well

• Joule to electron volt conversion $1J=6.24*10^{18}eV$

From the question, it is given that

the electron is in the fifth excited state that is $n=6$

the width of the potential well is $L=1*10^{-10}m$

we know that mass of an electron $m=9.1*10^{-31} kg$

substitute these values to get energy as follows

$E_{2} =\frac{6^{2} (6.626*10^{-34} )^{2} }{8*9.1*10^{-31} (1*10^{-10}) ^{2} }=21.67*10^{-17}J$

The energy in eV is $21.67*10^{-17}*6.24*10^{18}=1352eV$