Physics from school

Excerpt from a School Physics test in 2010 (Germany)

(11th/12th Grade = 16-17 years of age)*

Assignment 1:

For the measurement of the atomic mass of unknown particles, the depicted representation is used: A beam of unknown ions first traverses the crossed fields E and B0 in a straight line to be directed to a scanner after passing through an optical system in the field B1.

a) How is it possible for charged particles to pass through the capacitor without distraction? Determine the speed of the particles with your explanation. (E = 10³ V/m, B0 = 20 mT)

b) The particle beam considered consists of simply ionized atoms of an isotopic mixture of ¹²C, ¹³C and ¹⁴C. These isotopes have these atomic masses: 2,01*10^-26 kg, 2,17*10^-26 kg and 2,34*10^-26 kg. Where do they collide with the scanner? It is B1 = 90 mT.

Assignment 2:

Silver stands in the periodic table as ^107,87₄₇ Ag. The mechanical density is ρ = 10,5 g / cm³. Calculate the charge carrier density assuming that each silver atom submits 1 electron to the metallic grid.

1 a)
It is possible that charged particles pass through the capacitor without deflection, as the ions pass vertically through the electric field, which is also interspersed perpendicularly with a magnetic field. The electric field causes the force F_(el.) = Q*E to act on the ions and the opposite force F_(magn.) = Q*V*B because of the magnetic field.
As a result, the capacitor acts as a speed filter. So Q*E = Q*V*B
⇒ V = E / B with:
E → V*m^-1 = kg*m*s^-3*A^-1 and B → T = V*s*m^-2 = kg*s^-2*A^-1
So the velocity of the individual particles are the same and not dependent on the mass, but only on the electric and magnetic field.
⇒ V = ( 10³ / m ) / 20 mT
⇔ 5*10⁴ m/s

1 b)
Here, the Lorentz force acts as a centripetal force, whereby the ions describe a circular motion.
Therefore, the isotopes of different atomic masses occur at different locations on the scanner.
So: Q*V*B = ( m*V² ) / r
⇔ r = (m*V) / (Q*B) with:
Q as the elementary charge q_e = 1,6022*10^-19 C and C → As
⇒ r(¹²C) = (2,01*10^-26 kg * (5*10⁴) m/s) / (1,6022*10^-19 C * 90 mT)
⇒ r(¹²C) = 6,9697*10^-5 m

⇒ r(¹³C) = 7,5245*10^-5 m
⇒ r(¹⁴C) = 8,114*10^-5 m

2
density = (molar mass / Avogadro constant(N_A)) / V_atom
⇒ 10,5 g/cm³ = (107,87 g / 6,0221*10²³ mol^-1) / V_atom
⇔ V_atom = 1,705*10^-23 cm^-3

⇒ 1 / 1,705*10^-23 cm³ = 5,87*10²² cm^-3

Excerpt school-physics test / SVG play around