1. Path looks like parabolic and not circular. It goes with the field lines so had to be positive.
2. direct application of fromula for de Broglie wavelength where it is h divide by momentum.
3.The experiment does not tell me how the nucleus is arranged. Hence it it not structure of nucleus but size of nucleus only.
4. Current can tell me number of ions per s if I divide current by charge . If the number of alpha is known then I can find number of ions produced by one alpha. Number per s = No of ions per s divide by number of alpha per s .
5. Magnetic force = centripetal force.
b) the ratio can be obtained by just considering mass and charge. The charge of alpha is 2X the charge of beta. The mass of alpha is 4u and the electron is 9.11x 10-31 kg. Answer is 3640.
c) The radius of alpha is 25.9 m. Subs. the correct mass of alpha (4u) and the charge is 2e.
The radius of electron is 7.12 x 10-3 m. The mass of electron is 9.11 x 10-31 and charge is e.
d ) Draw the alpha path of large radius (nearly straight) out of the vertical face of box. Label A.
Draw eelctron path of 0.71 cm radius , hence exit the bottom horizontal face of box. Label B.
For both path draw straight line once it exit the box.
6. a) Define ...
b) Gravitational force is attractive in nature. Hence a system does work by its own field to move closer to the source (mass) . By def. of potential which is external work done must therefore assume negative values.
c) The change in gravitationa potential is the difference of the values of potential at the two location , surface and at an altitude.
ii) the speed of the projection to reach that alitude must therefore have KE to overcome the increase in PE. Hence m(change in potential) = 1/2 mv2. Cancel m , v can be found. Many mistaken the speed to be escape speed which is projection into outer space.
d) the acceleration is not a constant around the Earth and hence the formula given cannot be used .
7. Draw the resultant graph having the same freq as the first graph but having wavy edges due to wave 2. Wave 2 have small amplitude and hence does not change the freq. This is in fact what we see in resonance tube where the fundamental freq is dominant and overtones are produced to enhance the sound of the musical instrument.
b) Draw equal spacing. For large gap the straight lines passes through and only curve at its edges. The small gap produces circular waves. Draw two lines to show how it spread out. The big gap spread out with small angle but the small gap spread out nearly 180 degrees.
c) the length is calculated in terms of number of wavelengths. Hence the pd difference is in terms of number of wavelengths. 33.3 - 30.8 = 2.5
Hence for iii) the intensity is zero (minima) when the two waves meet with p.d 2.5 wavelength which is out of phase.
As the dectector moves from P to O , the detector receives 3 maxima including the maxima at O. You can deduce the maxima by counting from O , n=1 (M) n=2(M) again then n=2.5 is not a maxima . Not wrong to count minima . P is minima , as the detector moves towards P it will encouter n=1.5 (m) and n=0.5 (m). Hence 2 minima only from P to O.
8. a)The equation shows energy released hence Y must have higher binding energy than Sr.
b) def of decay constant
c) i) the decay constant must be in s-1 as the activity is in Bq.
ii) the mass can be calculated either using the Avogadro constant and molar mass formula or mass = N ( 90u) as each atom is approx 90 nucleons.
iii) the ratio is 0.882 . Can leave the decay constant in yr-1 as time is 5 yrs.
