Consider an electron inside the potential well of dimension 8.9 nm wide. The ground state energy of this electron is closest to: а. О.0056 eV O b. 0.0046 eV C. 0.0066 eV d. 0.0086 eV
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- Suppose you recently discovered a hydrogen like element that has only one electron orbiting around a nucleus containing a proton and a neutron. You found the ground state energy of the electron to be -16 eV. What will be the energy of this electron when it is on the excited state shown in the sketch? Note that all other possible intermediate states are shown by dashed lines. Electron is here Ground state 1.0 eV 16 eV - 1.0 eV -4.0 eV 4.0 eV3. This problem to be handed in for grading by 11:59 pm, Fri. April 10th The radial probability density, P(r), is defined such that the probability to find the electron between spheres of radius r and r+ dr is given by P(r)dr. In spherical coordinates with a radially symmetric potential P(r) = p2 Rn1(r)|* . Here the r² term is present because we are working in spherical coordinates. See sections 7.3 and 7.4 of Krane. (a) For an electron in the 2s state of the hydrogen atom, find the probability to find the electron inside the proton. Assume that the proton is a sphere of radius R = 8.414 x 1016 m. Show your work. Radial wave functions can be found in Table 7.1 of Krane. (b) For an electron in the 2p state of the hydrogen atom, find the probability to find the electron inside the proton. Assume that the proton is a sphere of radius R 8.414 x 10-16 m. Show your work. (c) Evaluate these two probabilities for muonic hydrogen. 12An electron in an excited energy state of the Hydrogen atom has an energy En = 0.85 eV which is 12.75 eV above the ground state. What is the radius rn of the electron’s orbit? Select one: a. 4.77 angstroms b. 8.47 angstroms c. 13.23 angstroms d. 2.12 angstroms