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Step 4 of 6 Calculate the energy for an electron in a 0.43-nm box using the formula from Step 3. E = 490.26 X 840 kJ-mol-1 Step 5 of 6 Calculate the exact energy for an electron in the first excited state in a 0.43-nm box. Recall that for a particle in a one-dimensional box we can write En = Eexact=49197 x KJ-mol-1 Submit n²h² we can therefore calculate an exact solution. 8mL 2'

Step 4 of 6 Calculate the energy for an electron in a 0.43-nm box using the formula from Step 3. E = 490.26 X 840 kJ-mol-1 Step 5 of 6 Calculate the exact energy for an electron in the first excited state in a 0.43-nm box. Recall that for a particle in a one-dimensional box we can write En = Eexact=49197 x KJ-mol-1 Submit n²h² we can therefore calculate an exact solution. 8mL 2'

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15 6x4 – 12Lx³ + 15 L²x? - 3,3 (- L²r? L3x 6r4 - N2 2me 12La3 dx = dx 2 2 2 Evaluate the integral in the numerator. p*H@ dx = N² 40m h² L³ e 40me The denominator of the E expression from Step 1 is 6 _ 3Lx + 13 ,2 4 -L²x 4 Lx³ + F8-3Lz' + 12rt - (%) Lz3 + +L교2 p dx = N² dx Evaluate the integral in the denominator. 1 1 φ φ αχx = N2 840 840 Step 3 of 6 Divide the numerator by the denominator (both from Step 2) and simplify. (Use the following as necessary: ħ, L, me, P, T, and x.) 21h? E L²m 21h? L²me e Step 4 of 6 Calculate the energy for an electron in a 0.43-nm box using the formula from Step 3. = 4.0 0.26 X 840 kJ•mol-1 φ Step 5 of 6 Calculate the exact energy for an electron in the first excited state in a 0.43-nm box. n2h2 Recall that for a particle in a one-dimensional box we can write En we can therefore calculate an exact solution. 8mL2' Eexact = 4.0 197 X kJ-mol-1 Submit
Transcribed Image Text:15 6x4 – 12Lx³ + 15 L²x? - 3,3 (- L²r? L3x 6r4 - N2 2me 12La3 dx = dx 2 2 2 Evaluate the integral in the numerator. p*H@ dx = N² 40m h² L³ e 40me The denominator of the E expression from Step 1 is 6 _ 3Lx + 13 ,2 4 -L²x 4 Lx³ + F8-3Lz' + 12rt - (%) Lz3 + +L교2 p dx = N² dx Evaluate the integral in the denominator. 1 1 φ φ αχx = N2 840 840 Step 3 of 6 Divide the numerator by the denominator (both from Step 2) and simplify. (Use the following as necessary: ħ, L, me, P, T, and x.) 21h? E L²m 21h? L²me e Step 4 of 6 Calculate the energy for an electron in a 0.43-nm box using the formula from Step 3. = 4.0 0.26 X 840 kJ•mol-1 φ Step 5 of 6 Calculate the exact energy for an electron in the first excited state in a 0.43-nm box. n2h2 Recall that for a particle in a one-dimensional box we can write En we can therefore calculate an exact solution. 8mL2' Eexact = 4.0 197 X kJ-mol-1 Submit
The variational method generally applies to ground-state wave functions. It may be extended to excited states under the condition that the trial wave function is orthogonal to the exact ground-state wave function. Under these circumstances, the trial wave function gives an upper limit for the energy of the first excited state. Given that the wave function, P(x) = M(x³ - Lx² + 3x) 2 where N is a normalization constant, satisfies these conditions for a particle in a box of length L, determine the formula for the variational energy of the lowest-energy excited state. Next, calculate the energy for an electron in a 0.43-nm box. Then calculate the exact energy for an electron in the first excited state in a 0.43-nm box. How good an approximation is this trial wave function? Step 1 of 6 The variational principle expresses Eo as the following (where p is the trial wave function). P"P dx O E P dx x*x dp O E = x*Âx dp o*¤ dx dx x*Hx dº O E x*x dp Step 2 of 6 Determine the numerator and denominator of the E expression from Step 1. (Use the following as necessary: ħ, L, me, P, TT, and x.) Since p(x) = N(x3 -x + x), the numerator of the E, expression from Step 1 is 2 2 3 15 3. 6x – 12LX³ + 15 L 6x4 – 12La³ + Ľ²r² - L3x P N2 2me dx 2 -) - Evaluate the integral in the numerator. 2-5 p*#@ dx = N² 40m 40m, e
Transcribed Image Text:The variational method generally applies to ground-state wave functions. It may be extended to excited states under the condition that the trial wave function is orthogonal to the exact ground-state wave function. Under these circumstances, the trial wave function gives an upper limit for the energy of the first excited state. Given that the wave function, P(x) = M(x³ - Lx² + 3x) 2 where N is a normalization constant, satisfies these conditions for a particle in a box of length L, determine the formula for the variational energy of the lowest-energy excited state. Next, calculate the energy for an electron in a 0.43-nm box. Then calculate the exact energy for an electron in the first excited state in a 0.43-nm box. How good an approximation is this trial wave function? Step 1 of 6 The variational principle expresses Eo as the following (where p is the trial wave function). P"P dx O E P dx x*x dp O E = x*Âx dp o*¤ dx dx x*Hx dº O E x*x dp Step 2 of 6 Determine the numerator and denominator of the E expression from Step 1. (Use the following as necessary: ħ, L, me, P, TT, and x.) Since p(x) = N(x3 -x + x), the numerator of the E, expression from Step 1 is 2 2 3 15 3. 6x – 12LX³ + 15 L 6x4 – 12La³ + Ľ²r² - L3x P N2 2me dx 2 -) - Evaluate the integral in the numerator. 2-5 p*#@ dx = N² 40m 40m, e
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