! 1. For an electron in a one-dimensional box of length 10 Å, calculate the following:a) The zero point energy in units of J, eV, and cm.b) The probability of finding the electron within a distance dL of the center of the box for thefirst three energy states.So I got a which is 6.023*1O^-34 then converted them to ev and cm-15A is the middle of the box so x is 5 in the bottom equation.I have the equation sqrt(2/L)sin(n(pi)x/L) so at n =11 got 1225 but it's wrong apparently it's supposed to be 2*10^9 but i dont know how to getthat through the equation.

Solution: a). The energy eigenvalue of an electron in one dimension box of length L is given by the…

1. For an electron in a one-dimensional box of length 10 Å, calculate the following: a) The zero point energy in units of J, eV, and cm'. b) The probability of finding the electron within a distance dL of the center of the box for the first three energy states. Sol got a which is 6.023*10^-34 then converted them to ev and cm-1 5A is the middle of the box so x is 5 in the bottom equation. have the equation sqrt(2/L)sin(n(pi)x/L) so at n =11 got 1225 but it's wrong apparently it's supposed to be 2*10^9 but i dont know how to get hat through the equation.

1. For an electron in a one-dimensional box of length 10 Å, calculate the following: a) The zero point energy in units of J, eV, and cm'. b) The probability of finding the electron within a distance dL of the center of the box for the first three energy states. Sol got a which is 6.02310^-34 then converted them to ev and cm-1 5A is the middle of the box so x is 5 in the bottom equation. have the equation sqrt(2/L)sin(n(pi)x/L) so at n =11 got 1225 but it's wrong apparently it's supposed to be 210^9 but i dont know how to get hat through the equation.