- #1
alexlee33
- 8
- 0
I have a take home quiz here and it would be greatly appreciated if someone could check over my work (it's numeric response btw):
1) A singly charged lithium ion (m= 1.16x10^-26 kg) is accelerated from rest thru a potential difference of 5.60 x 10^5 V. The velocity selector consists of an electric field (|E|= 4.98 x 10^5 V/m) and a magnetic field. The lithium ion finally enters a magnetic field (B= 0.650 T) perpendicular to it in the ion separation region.
a) What is the energy of the ion?
Ep = Ek
Ep = qV
Ep = (1.6 x 10^-19)(5.6 x 10^5)
FINAL ANSWER: 8.96 x 10^-14
b) What is the magnitude of the magn. field in the velocity selector if the ion remains undeflected?
Fe = Fm
q|E| = qvB
|E| / v = B
To find velocity...
qV = 0.5mv^2
v = (square root)((2(1.6x10^-19)(5.6x10^5) / (1.16x10^-26))
v = 3.93 x 10^6 m/s
Back to |E| / v = B...
(4.98x10^5) / (3.93x10^6) = B
FINAL ANSWER: 0.127 T
c) What is the lithium's radius of deflection in the ion separation region?
r = mv / qB
r = ((1.16x10^-26)(3.93x10^6)) / ((1.6x10^-19)(.65))
FINAL ANSWER: 0.438 m
- - - - - - - - - - - - - - - - - - - - - - - - - - - -
2. In a photoelectric cell the minimum voltage req. to reduce the current thru the cell to zero is 8.3 volts. Whats the max speed of the electrons ejected from the photoelectric surface in this cell?
Ep = Ek
qV = 0.5mv^2
v = (square root)((2(1.6x10^-19)(8.3) / (9,11x10^-31))
FINAL ANSWER: 1.7 x 10^6 m/s
- - - - - - - - - - - - - - - - - - - - - - - - - - - -
3. X-rays with a min. wavelength of 5.12 x 10^-10 m are produced by an xray tube. Whats the pot. diff. used in operating this tube?
qV = hc / (wavelength)
V = hc / q(wavelength)
V = ((6.63x10^-34)(3x10^8)) / ((1.6x10^-19)(5.12x10^-10))
FINAL ANSWER: 2.43 x 10^3 V
- - - - - - - - - - - - - - - - - - - - - - - - - - - -
4. If the 2nd energy level of a hydrogen-like atom is -2.44 x 10^-18 J, what is the energy of the 4th level?
En = E1 / n^2
Find E1... blah blah... do equation again using found E1...
FINAL ANSWER: -6.10 x 10^-19 J
-------
Thanks! :D
1) A singly charged lithium ion (m= 1.16x10^-26 kg) is accelerated from rest thru a potential difference of 5.60 x 10^5 V. The velocity selector consists of an electric field (|E|= 4.98 x 10^5 V/m) and a magnetic field. The lithium ion finally enters a magnetic field (B= 0.650 T) perpendicular to it in the ion separation region.
a) What is the energy of the ion?
Ep = Ek
Ep = qV
Ep = (1.6 x 10^-19)(5.6 x 10^5)
FINAL ANSWER: 8.96 x 10^-14
b) What is the magnitude of the magn. field in the velocity selector if the ion remains undeflected?
Fe = Fm
q|E| = qvB
|E| / v = B
To find velocity...
qV = 0.5mv^2
v = (square root)((2(1.6x10^-19)(5.6x10^5) / (1.16x10^-26))
v = 3.93 x 10^6 m/s
Back to |E| / v = B...
(4.98x10^5) / (3.93x10^6) = B
FINAL ANSWER: 0.127 T
c) What is the lithium's radius of deflection in the ion separation region?
r = mv / qB
r = ((1.16x10^-26)(3.93x10^6)) / ((1.6x10^-19)(.65))
FINAL ANSWER: 0.438 m
- - - - - - - - - - - - - - - - - - - - - - - - - - - -
2. In a photoelectric cell the minimum voltage req. to reduce the current thru the cell to zero is 8.3 volts. Whats the max speed of the electrons ejected from the photoelectric surface in this cell?
Ep = Ek
qV = 0.5mv^2
v = (square root)((2(1.6x10^-19)(8.3) / (9,11x10^-31))
FINAL ANSWER: 1.7 x 10^6 m/s
- - - - - - - - - - - - - - - - - - - - - - - - - - - -
3. X-rays with a min. wavelength of 5.12 x 10^-10 m are produced by an xray tube. Whats the pot. diff. used in operating this tube?
qV = hc / (wavelength)
V = hc / q(wavelength)
V = ((6.63x10^-34)(3x10^8)) / ((1.6x10^-19)(5.12x10^-10))
FINAL ANSWER: 2.43 x 10^3 V
- - - - - - - - - - - - - - - - - - - - - - - - - - - -
4. If the 2nd energy level of a hydrogen-like atom is -2.44 x 10^-18 J, what is the energy of the 4th level?
En = E1 / n^2
Find E1... blah blah... do equation again using found E1...
FINAL ANSWER: -6.10 x 10^-19 J
-------
Thanks! :D