(a)
The value of
(a)
Answer to Problem 39P
The value of
Explanation of Solution
Given:
The radius of Bohr bolt is
Formula used:
The expression for the spherical wave function for
The expression for the constant term
The new expression for the spherical wave function is given as,
Calculation:
The atomic number of hydrogen atom is
The spherical wave function for
Conclusion:
Therefore, the value of
(b)
The value of
(b)
Answer to Problem 39P
The value of
Explanation of Solution
Given:
The radius of Bohr bolt is
Formula used:
The expression for the spherical wave function for
The expression for the constant term
The new expression for the spherical wave function is given as,
Calculation:
The atomic number of hydrogen atom is
The spherical wave function for
Conclusion:
Therefore, the value of
(c)
The value of radial probability density
(c)
Answer to Problem 39P
The value of radial probability density
Explanation of Solution
Given:
The radius of Bohr bolt is
Formula used:
The expression for the radial probability density of finding a particle at a position
Calculation:
The radial probability density is calculated as,
Conclusion:
Therefore, the value of radial probability density
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Chapter 36 Solutions
Physics for Scientists and Engineers
- 0, f(x) = -Tarrow_forwardConsidering the Bohr’s model, given that an electron is initially located at the ground state (n=1n=1) and it absorbs energy to jump to a particular energy level (n=nxn=nx). If the difference of the radius between the new energy level and the ground state is rnx−r1=5.247×10−9rnx−r1=5.247×10−9, determine nxnx and calculate how much energy is absorbed by the electron to jump to n=nxn=nx from n=1n=1. A. nx=9nx=9; absorbed energy is 13.4321 eV B. nx=10nx=10; absorbed energy is 13.464 eV C. nx=8nx=8; absorbed energy is 13.3875 eV D. nx=20nx=20; absorbed energy is 13.566 eV E. nx=6nx=6; absorbed energy is 13.22 eV F. nx=2nx=2; absorbed energy is 10.2 eV G. nx=12nx=12; absorbed energy is 13.506 eV H. nx=7nx=7; absorbed energy is 13.322 eVarrow_forwardIf the speed of the electron in Example 19-4 were 7.3 * 105m>s,what would be the corresponding orbital radius?arrow_forwardQ#03. Show that the (hkl) plane is perpendicular to the [hkl] direction.arrow_forwardSo Determine the distance between the electron and proton in an atom if the potential energy ?U of the electron is 15.4 eV (electronvolt, 1 eV =1.6×10−19=1.6×10−19 J). Give your answer in Angstrom (1 A = 10-10 m)arrow_forwardA box contains 10 indentical atoms with the following velocities. Find its Vrms atom 1 100 m/s atom 2 150 m/s atom 3 250 m/s atom 4 200 m/s atom 5 290 m/s atom 6 190 m/s atom 7 |130 m/s atom 8 230 m/s atom 9 160 m/s atom 10 260 m/s 204 m/s O 212 m/s 250 m/s O 267 m/sarrow_forwardChapter 4- questn-23 The position of an electron is given by i = 3.0tî – 4.0t2j + 2.0k , where r is in meter, t is in second. What is the angle between i and +x axis at t=2s. Option 1 Option 2 Option 3 Option 4 Option 5 78.4° 79.4° 80.4° 81.4° 82.4°arrow_forwardA mercury atom emits light at many wavelengths, two of which are at (1) 404.7 nm and (2) 435.8 nm . Both of these transitions are to the same final state. Determine the energy of each emitted wavelength (in J). A E1 = 7.908 x 10-19 J and E2 = 6.558 x10-19 J E1 = 4.628 x 10-19 J and E2 = 4.968 x10-19 J E1 = 5.908 x 10-19 J and E2 = 5.558 x10-19J E1 = 4.908 x 10-19 J and E2 = 4.558 x10-19 Jarrow_forward1. (a) Calculate the divergence of ā = x'â +2y' zŷ +3zx2. . 2 8 2yZY+ Le t x x xe=A 3zx2 2 1 3 2 = 2x :+4yZ + 3x 4 5 8/8 5 3 T 26 40 (b) Calculate the curl of b = x y & + 2y zŷ + 3zx2. Vx Z = ze 32X 2yZ 2yZ- XY -(0-2)2-(32- 0) 4 +(0 - x)2 =62y1-62)9-6)Źarrow_forwardThe position of an electron is given by = 3.0tî – 4.0t2+2.0k, wherer is in meter, t is in second. What is the angle between i and +x axis at t=2s. Option 1 Option 2 Option 3 Option 4 Option 5 78.4° 79.4 80.4° 81.4° 82.4°arrow_forwardUse the table to determine the energy in eV of the photon emitted when an electron jumps down from the n = 2 orbit to the n = 1 orbit of a hydrogen atom. Allowed Values of the Hydrogen Electron's Radius and Energy for Low Values of n n rn En 1 0.053 nm −13.60 eV 2 0.212 nm −3.40 eV 3 0.477 nm −1.51 eV 4 0.848 nm −0.85 eV eVarrow_forwardn = 1 E = -13.6 ev n= 4 E = -0.85 ev n=5 n= 2 E = -0.54 ev E = -3.4 ev n= 3 E = -1.51 ev a) At what wavelength (in meters) does the n = 3 → 2 transition of hydrogen occur?arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning