Correct answer 2472 89 nm explanation let m e 9 kg h 1

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.Findthewavelengthofthefirstmem-berofthisseries.En=n2π2¯h22m L2,soΔE=E2-E1=(22-12)π2¯h22m L2=3π2(1.05457×1034J·s)22 (9.10939×1031kg)×1(1.5×109m)2= 8.0329×1020J.Since ΔE=h f,f=E2-E12π¯h=8.0329×1020J2π(1.05457×1034J·s)= 1.21232×1014s1andλ=cf=2.99792×108m/s1.21232×1014s1×109nm1 m=2472.89 nm.011(part1of2)10.0pointsIf, in1λ=Ryparenleftbigg1n21-1n22parenrightbigg,you setn1= 1 and taken2greater than 1,you generate what is known as theLymanseries.
sornia (vgs339) – Homework 10 – yao – (54785)5
Consider the next three members of this se-ries. The wavelengths of successive membersof the Lyman series approach a common limitasn2→ ∞.What is this limit?Which of the following photon energiesEγcould NOT be found in the emission spectraof this atom after it has been excited to then= 4 state?Correct answer: 91.1288 nm.Explanation:λ3=1(1.09735×107m1)parenleftbigg1-19parenrightbigg×109nmm= 102.52 nm,λ4=1(1.09735×107m1)parenleftbigg1-116parenrightbigg×109nmm= 97.204 nm,λ5=1(1.09735×107m1)parenleftbigg1-125parenrightbigg×109nmm= 94.9258 nm,...Eγ= 3 eV2.Eγ= 5 eV3.Eγ= 1 eV4.Eγ= 4 eVcorrect5.Eγ= 2 eVExplanation:By subtraction, the transition fromn= 4ton= 3 would release 1 eV; fromn= 4ton= 2 would release 3 eV; fromn= 3 ton= 2 would release 2 eV; fromn= 3 ton= 1would release 5 eV. The only choice remainingis4 eV.014(part2of2)10.0pointsWhich of the following transitions will pro-duce the photon with the longest wavelength?
1.Eγ= 3 eV2.Eγ= 5 eV3.Eγ= 1 eV4.Eγ= 4 eVcorrect5.Eγ= 2 eVExplanation:By subtraction, the transition fromn= 4ton= 3 would release 1 eV; fromn= 4ton= 2 would release 3 eV; fromn= 3 ton= 2 would release 2 eV; fromn= 3 ton= 1would release 5 eV. The only choice remainingis4 eV.014(part2of2)10.0pointsWhich of the following transitions will pro-duce the photon with the longest wavelength?
sornia (vgs339) – Homework 10 – yao – (54785)61.n= 2ton= 12.n= 3ton= 13.n= 4ton= 14.n= 4ton= 3correct5.n= 3ton= 2Explanation:Thelongestwavelengthisproducedinthe case where the least amount of energyis released.That is,in this case,fromn= 4 ton= 3.015(part1of2)10.0pointsHow much energy is required to cause an elec-tron in hydrogen to move from then= 3 stateto then= 6 state?Correct answer: 1.13333 eV.Explanation:Let :nf= 6,ni= 3,andk= 13.6 eV.The energy of thenstate isEn=-kn2.The energy absorbed will equal the energydifference between these two states.E=Enf-Eni=-kparenleftBigg1n2f-1n2iparenrightBigg= (-13.6 eV)parenleftbigg162-132parenrightbigg=1.13333 eV.016(part2of2)10.0pointsIf the electrons gain this energy by collisionbetween hydrogen atoms in a high tempera-ture gas, find the minimum temperature ofthe heated hydrogen gas. The thermal energy

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