The speed of an unknown particle is 0.95c and its momentum is 2855 MeV/c. WHICH
ID: 1483411 • Letter: T
Question
The speed of an unknown particle is 0.95c and its momentum is 2855 MeV/c. WHICH of the following is the unknown particle most likely to be? (hint, calculate the rest mass, M = 10^6, G = 10^9) An electron with rest mass 0.511 MeV/c^2 A muon with rest mass 105.7 MeV/c^2 A pion with rest mass 139.6 MeV/c^2 A proton with rest mass 938.3 MeV/c^2 An alpha particle with rest mass 3.727 GeV/c^2 Light of wavelength 225 nm falls on aluminum, which has a work function of 4.3 eV. What is the maximum KINETIC ENERGY (in eV) of the electrons that are ejected due to the photoelectric effect? An electron is accelerated from rest through a potential difference of 600 V. What is the de Broglie WAVELENGTH of the electron? (hint, the electron reaches a non-relativistic speed) 20 cm 0.6 cm 1.7 times10^34 m 6.2 times10^8 m 5.0 times 10^ 11 m The numbers below correspond to changes in the principle quantum number, n for the Bohr model of the hydrogen atom. For which electronic transition does the electron LOSE the most energy? (hint, you should be able to answer this with a quick sketch)Explanation / Answer
70.
Use equation,
Relativistic momentum , P= mu/sqrt(1-(v/c)^2)
u= 0.95c , P= 2855MeV/c
Plugging values,
2855MeV/c = m*0.95c/sqrt(1-0.95^2)
Gives, m=938.3 MeV/c^2
This is rest mass of proton.
Thus correct choice is : D
71.
Use equation,
KE(max) = hc/ - = (6.62*10^-34*3*10^8)/(225*10^-9) - (4.3*1.6*10^-19) = 1.94*10^-19 J = 1.2 eV
Correct choice is : A
72
By de Broglie wavelength,
=h/p=h/mv ------------------(1)
KE = 1/2mv^2 = qV =eV --------------(2)
From (1) and (2),
v=sqrt(2eV/m)
Plugging in (1),
= h/[m(sqrt(2eV/m)] = h/sqrt(2eVm) = (6.62*10^-34)/sqrt(2*1.6*10^-19*600*9.1*10^-31) = 5.0*10^-11 m
Correct choice is: E
73.
Use equation,
E = -13.6[1/n1^2 - 1/n2^2]eV
Gives max E for transition from n2=1 to n1= 3
Thus correct choice is: A 1 -> 3
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