Why Do Some Students Struggle with A-Level Physics Concepts?

[OJ-Hick]

Last year I sat my physics A-Level paper and despite achieving A's in the first 3 modules I took a nosedive and underachieved. I have finished school now and am retaking modules in January time and upon my preparation I have come across a few problems which I can't seem to answer! Any help would be greatly appreciated. Here goes:

1) Explain the problem and state how the measurement can be made more accurate when measuring the fringe width in Young's Double Slit experiment (laser/monochromatic light source)

2) Explain how the following formula follows from the definition of intensity:
I= P/(4 x pi x r2)

3) Photoelectical Emission:
(i)When p.d is set to zero the microammeter indicates a small current. Why?

(ii) The cathode is made from sodium - work function 3.7 x 10^-19 J. Show the stopping potential for radiation of frequency 7.5 x 10^14 Hz is approx 0.8V

(iii) Sketch a graph to show how current changes as the reverse potential difference V is increased from 0 to 1.0 V


THANKS GUYS HOPE YOU CAN HELP

 

[jbsteele]

!1) In Young's double slit experiment, the fringe width can be made more accurate by using a laser or monochromatic light source. This allows for the light to be more focused and concentrated, creating sharper fringes with a greater contrast between the bright and dark fringes. Additionally, a more powerful light source can be used to produce higher intensity of light across the entire screen, further increasing the accuracy of the measurement. 2) The formula I=P/(4 x pi x r2) follows from the definition of intensity, which is the amount of energy per unit area that is passing through a given surface. The formula can be derived by considering the total power being emitted from a point source, P, and the fact that this power will be spread out in all directions. As the distance, r, increases, the total area over which the power is spread increases, thus decreasing the intensity at any given point. This can be expressed mathematically through the equation I = P/(4 x pi x r2). 3) (i) When the p.d is set to zero, the microammeter indicates a small current because there is a small amount of current present due to thermal electrons. These electrons are generated from thermal energy within the cathode and are not related to the photoelectric effect. (ii) The stopping potential for radiation of frequency 7.5 x 10^14 Hz can be calculated using the equation V = hf/e, where h is Planck's constant (6.63 x 10^-34 Js), f is the frequency of the radiation, and e is the charge on an electron (1.60 x 10^-19 C). Substituting the values given into the equation yields V = 0.8 volts. (iii) A graph to show how current changes as the reverse potential difference V is increased from 0 to 1.0 V would look something like this: | | || * | | * || * || * || * ||* ||-----------------------| 0 1.0V

 

[kerimek]

1) The problem in the Young's Double Slit experiment is that the measurement of the fringe width can be affected by factors such as the accuracy of the laser/monochromatic light source and the positioning of the slits. To make the measurement more accurate, one can use a more precise laser or light source, and ensure that the slits are properly aligned and the distance between them is accurately measured. Another way to improve accuracy is to increase the distance between the slits and the screen, which will result in a larger fringe width and therefore easier to measure.

2) The formula for intensity follows from the definition of intensity as the power per unit area. The power is represented by the symbol P, and the area is represented by 4 x pi x r2, where r is the radius of the circle. This formula is derived from the fact that the intensity of light decreases as the distance from the source increases, and the area of a circle increases with the square of the radius.

3) (i) When the p.d is set to zero, there is still a small current because the photoelectric effect is still occurring. Electrons are being emitted from the cathode due to the incident radiation, even though the potential difference is zero.

(ii) The stopping potential is the minimum potential difference needed to stop the photoelectric current. In this case, the frequency of the radiation is given as 7.5 x 10^14 Hz. Using the equation for the stopping potential, V = hf/W, where h is Planck's constant and W is the work function, we can calculate the stopping potential to be approximately 0.8V.

(iii) The graph of current vs reverse potential difference V would show a linear increase in current as V increases from 0 to the stopping potential, and then a plateau at zero current for higher values of V. This is because at lower values of V, the electrons have enough energy to overcome the work function and be emitted from the cathode, resulting in an increase in current. However, once the stopping potential is reached, no more electrons can be emitted and the current remains at zero.