Consider the following energy levels of a hypothetical atom: E4 _____ -1.0 times

Question

Consider the following energy levels of a hypothetical atom: E4 _____ -1.0 times 10-29 J E3 _____ -5.0 times 10-29 J E2 _____ -10 times 10-29 J E1 _____ -15 times 10-29 J (a) What is the wavelength of the photon needed to excite an electron from E1 to E4? (b) What is the energy (in joules) a photon must have in order to excite an electron from E2 to E3? (c) When an electron drops from the E1 level to the E1 level, the atom is said to undergo emission. Calculate the wavelength of the photon emitted in this process. Give the value of the four quantum numbers of an electron in the following orbitals: (a)3s, (b)4p, (c)3d. The ground-state electron configuration listed here are incorrect. Explain what mistakes have been made in each and write the correct electron configurations. A1: 1s2 2s2 2p4 3s2 3p2 B: 1s2 2s2 2p2 F: 1s2 2s2 2p6 Write the complete and abbreviated ground-state electron configuration for the following elements: B, V, Ni, As, I, Au.

Explanation / Answer

3.
a. DE= E4-E2
= -1*10^-19-(-10*10^-19) = 9*10^-19 j
lambda = hc/DE
= 6.626*10^-34*3*10^8/9*10^-19
Wavelength = 2.2*10^-7 m

b. DE= E3-E2 = (-5*10^-19)-(-10*10^-19)
= 5*10^-19 j

c. DE= E1-E3
= -15*10^-19-(-5*10^-19) = -10*10^-19 j
lambda = hc/DE
= 6.626*10^-34*3*10^8/10*10^-19
emission light wavelength = 1.99*10^-5 m

4.
a. 3s n=3,l=0,ml=0,s=+1/2
b. 4p n=4,l=1,ml=-1,s=+1/2
c.3d n=3,l=2,ml=-2,s=+1/2

5.
Correct configuration is
Al= 13 from the given configuration 2p is in complete . but electrons were filled into 3rd cell.so that its wrong. correct one is =1S2 2S2 2P6 3S2 3P1

B= 5 = given configuration 1s2 2s2 2p5 but boron have only 5 electrons.so that its wrong the correct one is 1S2 2S2 2P1
F=9 = given configuration 1s2 2s2 2p6 but flourine have only 9 electrons.so that its wrong the correct one is 1S2 2S2 2P5

6.
B= 5 = 1S2 2S2 2P1
V= 23 = 1s2 2s2 2p6 3s2 3p6 3d3 4s2   
Ni= 28 = 1s2 2s2 2p63s2   3p6 3d8 4s2
As = 33 = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3   
I = 53 = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p5   
Au = 79 = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 6s1

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