b) i) Planck's constant = gradient of line
m = change in y / change in x
m= 1.6 x 10^-19 / (7.4 x 10^14 - 4.9 x 10^14)
m= 6.4 x 10^-34 Js
Close to Planck's constant 6.6 x 10 ^-34
KEmax = hf -φ
y = mx + c
( - because its a negative y-intercept)
Finding φ at 1.6 x 10^-19 and 7.4 x 10 ^14 Hz,
6.6 x 10 ^-34 x 7.4 x 10 ^14 = φ + 1.6 x 10^-19
φ= 3.284 x 10^-19
= 3.2 x 10^-19
b) If the frequency is lower than threshold frequency then KEmax will be more than hf and so subtracting KEmax from hf would yield a negative value. The work function cannot be a negative value.
2) a) Changes that occur in the micrometer when...
i) The intensity of incident light is increased (but frequency remains the same)
An increase in intensity will mean that more electrons will be liberated. As Q= IT, more electrons will mean more charge per unit time (Q/T increases) and as I = Q/T, I (current) will also increase. The microammeter will therefore detect an increase in current.
ii) The effect of increasing frequency will depend as to whether the starting frequency is the threshold frequency or not. If it is and the frequency is increased above the threshold frequency, the maximum energy of the electrons will depend on the frequency of the incident light and will be higher as the frequency gets higher. If it is below threshold frequency and is increased to a value still below the threshold frequency, no photoelectrons will be emitted the current is still zero.
b) i)
ii) Ek = 1.9 x 1.6 x 10^-19 = 3.04 x 10^-19 J
540 nm = 540 x 10^-9 m
v= fλ
3 x 10^8 = 540 x 10^-9 x f
f= 5.56 x 10^14 Hz
φ = hf - KEmax
φ= (6.6 x 10 ^-34 x 5.56x10^14) - 3.04 x 10^-19
φ= 6.3 x 10^-20
3) a) The de Brogile wavelength of a particle is the wavelength of a wave that determines the probability of a particle's position.
b)
5.0 kV= 5.0 x 10^3 V
5.0 x 10^3 x 1.6 x 10^-19= 8 x 10^-16 J
Ek = 1/2 x m x v^2
8 x 10^-16 = 1/2 x 9.11 x 10^-31 x v^2
v= 4.2 x 10^7 ms^-1
p=mv
=9.11x 10^-31 x 4.2 x 10^7
p=3.8 x 10^-23
λ = h / p
λ= 6.6 x 10 ^-34 / 3.8 x 10^-23
λ= 1.7 x 10^-11
4) a) V= 75 V
Ek = 75 x 1.6 x 10^-19 = 1.2 x 10 ^-17
Ek = 1/2 x m x v^2
1.2 x 10 ^-17= 1/2 x 9.11 x 10^-31 x v^2
v= 5.1 x 10^6 ms^-1
p=mv
p= 9.11x10^-31 x 5.1 x 10^6
p= 4.67 x 10 ^-24
λ = h / p
λ = 6.6 x 10 ^-34 / 4.67 x 10 ^-24
λ = 1.4 x 10^10 m
b) The general shape of the graph mirrors the shape of a diffraction pattern as can be easily discerned from the maxima and minima. This graph therefore confirms de Brogile's hypothesis that electrons have both wave and particle properties.
No comments:
Post a Comment