Consider the one-dimensional potential box of width a = 4 A. a) For the ground s

Question

Consider the one-dimensional potential box of width a = 4 A. a) For the ground state and first excited state (n = 1 and 2), draw the wavefunctions in the box (like book) and calculate the probability of finding a particle between 1.5 and 2.5 A. b) Why is above probability much smaller for n = 2? c) Calculate the energies for the states and the transition frequencies (in cm^-1) between the ground state and the first two excited states. Assume that the particle has the mass of an electron. d) For the first excited state, calculate the expectation (average) values and .

Explanation / Answer

a.

b. for box length l = a = 4 Angstron

wavelength = 2a/n

speed = v

so frequency, f = nv/2a

where n is the excited state number (1 for ground state, 2 for 1st excited and so on)

so, the wavefunction can be written as

phi(x) = Asin(2*pi*n*x/2*a) = Asin(n*pi*x/a)

so probability for finding an electron between x = 0.375a (1.5A) and 0.625a ( 2.5 A) is

P = integrate(phi^2(x)dx)/integrate(dx) [ from x1 = 0.375a to x2 = 0.625a]

P = integrate(A^2*sin^2(n(pi(x/a)dx)/(x2 - x1) = integrate(A^2*(1 - cos(2n*pi*x/a))dx)/2(x2 - x1)

P = A^2*(x - a*sin(2n*pi*x/a)/2n*pi)/2(x2 - x1)

if we integrate from x = 0 to x = a, P = 1

1 = A^2*(a - a*sin(2n*pi*a/a)/2n*pi - 0 + a*sin(2n*pi*0/a)/2n*pi)/2(a - 0)

1 = A^2/2

A = sqroot(2)

so, P = P = 4(0.625a - a*sin(2n*pi*0.625a/a)/2n*pi - 0.375a + a*sin(2n*pi* 0.375a/a)/2n*pi)/a

P = 4(0.25 - (sin(2n*pi*0.625) - sin(2n*pi* 0.375))/2n*pi)

P = 4(0.25 - (sin(n*pi*0.25)cos(n*pi))/n*pi)

so for n = 1

P = 0.95

for n = 2

P = 0.36

the probability is lesser for n = 2 as higher energy states are more difficult to achieve and hence leess probability of the particle being in higher energy state

c. energy = 0.5mv^2

but, lambda = h/mv

v = h/lambda*m

so, E = 0.5*m*h^2/lambda^2*m^2 = 0.5h^2/lambda^2*m

lambda = 2a/n

so, E = 0.5*n^2*h^2/4a^2*m

for n = 1

E = 0.5h^2/4a^2*m = 0.5(6.63*10^-34)^2/4*(4*10^-10)^2*9.1*10^-31 = 3.77337*10^-19 J = 2.358 eV

for n = 2

E = 9.4344 eV

k = 2*pi/lambda

k(n = 3) = 2*pi*3/2a = 3pi/a

k(n = 2) = 2*pi*2/2a = 2pi/a

k(n = 1) = 2*pi*/2a = pi/a

so from first to seciond excited state, k(2) - k(1) = pi/a = 78.53*10^6 per cm

so from first to third excited state, k(3) - k(1) = 2pi/a = 157.07*10^6 per cm

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.