Edexcel A-Level Physics Practical Analysis

Values of correct to 2 d.p. Accept 3 d.p.

Axes labelled: y as ( /^) and x as n

Accept degrees for ^

Appropriate scales chosen

Values plotted accurately

Best fit line drawn (1)

Using a large triangle on the best-fit line gives a gradient of about -0.0405. Since the gradient is $-\lambda$,

\lambdapprox0.041

(to 2 or 3 significant figures, with no unit).

From $\ln heta=\ln heta_0-\lambda n$,

$$\ln heta_0=\ln heta+\lambda n=3.2+(0.041)(50)=5.25,$$

so $heta_0=e^{5.25}=191^{\circ}$. This is greater than $180^{\circ}$, so the claim is correct.

Values of ln A correct and consistent to 3 d.p. (accept 2 d.p.).

Axes labelled: y as ln(A/cm) and x as n.

Appropriate scales chosen.

ln values plotted accurately.

Best fit line drawn.

ln(A/cm): 2.140, 1.946, 1.705, 1.609, 1.386, 1.253.

Uses large triangle to calculate gradient.

Value of λ in range 0.0335 to 0.0384.

Calculated value of λ given to 2 or 3 s.f., no unit, positive.

Example: λ = −gradient = −(2.10 − 1.23)/(5.5 − 30) = 0.0355.

Either there is significant scatter in the data, so the best fit line is uncertain; or there may be a systematic error in measurement of A, so the best fit line may be shifted; or the initial angle may be too large to apply the small angle approximation, so extrapolating back from a straight line is not valid.