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 6 Click here to jump to the problem! WoosterGrad2018-05-29 21:51:47 For the sake of other people studying for the GRE:\r\n\r\nC would be correct if the question asked for the expectation value of the energy-- the number you would get if you took many, many measurements and averaged them.\r\n\r\nHowever, the question asks for a possible value of ONE measurement. A single measurement of the energy will collapse the wave function to a single energy eigenstate.\r\n\r\nThe energy of a state is proportional to the square of the principle quantum number: $E_n=n^2E_1$.\r\n\r\nThe principle quantum numbers are 1, 2, and 3 in this superposition, so a single measurement would yield $E_1$, $4E_1$, or $9E_1$. \r\n\r\nThis reveals choice D as the correct answer. Click here to jump to the problem!
 11 Click here to jump to the problem! BillNyeTheRussianSpy2018-05-22 03:06:49 I think, notationally, it is easier to have $E_{n}=n^2 E_{1}$. But it still comes out the same.\r\n\r\n$E_{2}=2eV=4E_{1}$ implies $E_{1}=1/2eV$Click here to jump to the problem!
 12 Click here to jump to the problem! BillNyeTheRussianSpy2018-05-22 03:06:49 I think, notationally, it is easier to have $E_{n}=n^2 E_{1}$. But it still comes out the same.\r\n\r\n$E_{2}=2eV=4E_{1}$ implies $E_{1}=1/2eV$Click here to jump to the problem!
 14 Click here to jump to the problem! BillNyeTheRussianSpy2018-05-22 02:25:21 I just remembered $E=-\\mu \\cdot B$ from a class where $\\mu$ is the electron spin magnetic moment, not the nuclear magnetic momentClick here to jump to the problem!
 17 Click here to jump to the problem! timfinnigan2018-05-20 03:18:29 Plug $\\psi$ into the TDSE and then set x=0. You will end up with $E=(hb)^2 / 2M$ (should be h-bar). Put that into the TDSE and then cancel like terms. You will end up with V(x)=choice B.Click here to jump to the problem!
 18 Click here to jump to the problem! timfinnigan2018-05-19 03:40:44 Simple Work-KE Thm:\r\n\r\n$W=\\Delta KE=KE_{f}-KE_{i}=0-(1/2)(5)(10)^2 =-250$\r\n\r\n$W=Fd=F(.025)$\r\n\r\n$F=W/.025=-250/.025=-10^4$\r\n\r\nThe magnitude of the force, $10^4$, is the answer (the negative could be thought of as direction).Click here to jump to the problem!
 24 Click here to jump to the problem! TimToolMan2018-04-12 16:27:04 We know that $\\beta=v/c=pc/E$ (using relativistic momentum and energy). Since $v=c$ we have $1=pc/E$ which implies $E=pc$. And, from the relation $E^2 = (pc)^2+(mc^2)^2$, we can say that $E=pc$ implies $m=0$.Click here to jump to the problem!