01 Jan 01
14-6 = 8
the "absolute" largest e/m = electron: charge of e / very small
on electron is negative so really the "largest" considering sign of
charge = proton which is positive
it’s a bad question and I’m not sure which answer the question is after –
they’ll have to allow either answer so long as the reason is given.
elements with the same number of protons but different numbers of
work function = minimum energy required to remove an electron from the
Kinetic energy = the energy associated with the electron's emission
velocity if the incident radiation has more energy than the work function
as the intensity of the light is doubled then the current (the rate of
electrons emitted) will double for any frequency that is above the threshold -
sorry this is a change from my previous answer - since I have had a chance to
think about it - thanks go to Ben King for spotting this one
as the intensity of the light is doubled then the current (the rate of electrons emitted) will double for any frequency that is above the threshold - sorry this is a change from my previous answer - since I have had a chance to think about it - thanks go to Ben King for spotting this one
(ii) assuming that every photon absorbed results in the emission of one electron.
l = 350 x 10-9
c = f l
so f = c / l = 3 x 108
/ 350 x 10-9
f = 9 x 10 14 Hz
The threshold frequency is the minimum frequency required to emit electrons with zero kinetic energy. F
From Einstein's photoelectric equation hf = KE + Æ if we shine light with the threshold frequency then KE = zero therefore Einstein's equation reduces to hf = Æ
Æ = work function = hf = 6.63 x 10 -34
x 9 x 10 14
= 5.7 x 10 -19 J
experience the strong nuclear force - take part in the strong interaction
experience the weak nuclear force BUT NOT THE STRONG
b + antineutrino (use
neutrino ( I remember
this as start with a proton - end with a neutrino, but start with a neutron and
end with an antineutrino)
- the energy difference between one level and another is quantised - it
has a discrete value
- an excited electron can
lose energy by dropping from one energy level to another
- for a particular energy
level change that energy can be released as a quantum of light of a particular
frequency as frequency is proportional to energy
emission or capture of an electron by an atom - removal of the electron
in any energy level to infinity
22 x 10-19 J
energy required = 22 x 10-19 J
E = hf rearrange: f = E/h = 22 x 10-19 / 6.63 x 10 -34
= 3.3 x
10 15 Hz
it will be ionised with excess kinetic energy - so it will be emitted
with a velocity equivalent to the excess energy.
E = hf2 - hf3
= 5.4 - 2.4 (x 10-19)
3 x 10-19 J
E/h = 3 x 10-19 / 6.63 x 10 -34
= 4.52 x 10 14 Hz
rearrange: l = c / f =
3 x108 / 4.52 x 10 14
l = 6.64 x 10 -7m
zig zag line within the core reflecting so that i = r at the core,
n = 1 / sinC or
sinC = 1 /n = 1/ 1.5 = 0.66
therefore C = sin-1 0.66
C= 41.81 o [NB to 2 decimal places cos that’s the dps in the
(b) (i) cladding because: without cladding light rays would pass between different fibres at their point of contact.
(ii) the lower the refractive index of the cladding – the more likely it is that internal reflection will take place at the cladding core boundary – thus reducing the possibility of light leaking from the fibre. The refractive index of the core is 1.5, less than 1.45 reduces the tolerance for error.
(c) used to send information for cable tv, used to view the interior of the body during keyhole surgery etc …
6 (a) (i) Electron diffraction – electrons will diffract through crystals and diffraction is a property associated with waves.
(ii) the photoelectric effect – em radiation will emit electrons from a metal if exposed to radiation above the threshold frequency.
(iii) em radiation will refract – it bends at the boundary between optical media of different density.
(b) l = h / p = h / momentum = h / (mv) = 6.63 x 10-34 / (9.1 x 10-31 x 5 x 106)
= 1.45 x 10-7 m
7 (i) to reduce absorption of the alpha particles by collision with air molecules.
(ii) to reduce absorption of alpha particles – as alpha radiation is the least penetrating of all radiation and therefore ensuring that transmitted alpha particles will be detected.
(iii) most alpha particles travel through the gold foil undeflected
as the atom is mostly empty space,
some alpha particles are deflected through very large angles
which indicates that they are repelled by a very small but massive central nucleus.