Determine the radius of^238_92 U nucleus. A) 9.27 times 10^-15 m b) 7.39 times 1
ID: 1622270 • Letter: D
Question
Determine the radius of^238_92 U nucleus. A) 9.27 times 10^-15 m b) 7.39 times 10^-15 cm c) 7.39 times 10^-15 m d) 7.43 times 10^-15 m e) None Determine the volume of^238_92 U nucleus. A) 4.27 times 10^-42 m^3 b) 1.38 times 10^-42 cm^3 c) 2.34 times 10^-45 m^3 d) 5.35 times 10^-13 m e) None Calculate the density of^238_92 U nucleus. a) 1.27 times 10^44 kg/m^3 b) 1.38 times 10^44 kg/cm^3 c) 1.38 times 10^44 m^3 d) 6.38 times 10^44 m^3 e) None Calculate the density of^4_2 He nucleus. a) 1.27 times 10^-15 kg/m^3 b) 2.39 times 10^-15 kg/cm^3 c) 1.11 times 10^15 kg/m^3 d) 1.38 times 10^44 kg/m^3 e) None The electron of a hydrogen atom makes a transition from the n=5 state to the n=2 state. What is the frequency of the emitted photon? A) 344 MHz B) 234 MHz C) 2.34 times 10^14 Hz D) 3.48 times 10^14 Hz E) NoneExplanation / Answer
16 ans
Given that
atomic number Z=92
mass number A=238
now we find the radius
radius r=1.2*10^-15*238^1/3=7.3*10^-15
the correct option E
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17 ans
voume of nucleus 23892U=4/3*3.14*(7.3*10^-15)^3=1628.7*10^-45 m^3
the correct option is E
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18 ans
density of nucleus =mass/volume=238*1.66*10^-27/1628.7*10^-45
=0.233*10^18
=2.3*10^17 kg/m^3
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19 ans
density of nucleus 42He=A*1.66*10^-27/(4/3)*(1.2*10^-15)^3*A=
=2.3*10^17 kg/m^3
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20 ans
now we find the frequency of hydrogen atom n=5 to n=2
frequency =RZ^2[1/n1^2-1/n2^2]
=12.8*10^-18*2^2[1/2^2-1/5^2]
=51.2*10^-18[25-4/100]
=10.752*10^-18 Hz
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