2020 A/L Physics Past Paper · Question 01 (English Medium)

Answer: (5) $ML^2T^{-1}$

What the question asks

Write the dimensions of Planck's constant $h$ in $M$, $L$, $T$.

Then find the matching option.

Relevant theory

Symbolic expressions which show the relation of a physical quantity to basic units is called its dimension.

ΔE = E₂ − E₁ = hf

where h is the Planck constant and f the frequency of the absorbed photon.

We know: a photon's energy.

$$\Delta E = hf$$

So:

$$h = \dfrac{\Delta E}{f}$$

Energy is force times distance.

$$[E] = MLT^{-2} \times L = ML^2T^{-2}$$

Frequency is oscillations per second.

$$[f] = T^{-1}$$

$$[h] = \dfrac{ML^2T^{-2}}{T^{-1}} = ML^2T^{-1}$$

Reading the options

Each option's $M$, $L$, $T$ exponents are compared with $[h]$'s exponents $(1, 2, -1)$.

asked

M L^2 T^{-1}

$h = E / f$

1

M^2 L T

$M^2 L T$

2

M^2 L T^{-1}

$M^2 L T^{-1}$

3

M L T^2

$M L T^2$

4

M L T^{-1}

$M L T^{-1}$

5

$M L^2 T^{-1}$

Option (1) has $M^2$. This is wrong.

Option (2) has $M^2$. This is wrong.

Option (3) has $T^{2}$. We want $T^{-1}$. This is wrong.

Option (4) has $L^1$. We want $L^2$. This is wrong.

Only option (5) is left.

Checks

A quick check. Dimension check:

$$[h][f] = ML^2T^{-1} \times T^{-1} = ML^2T^{-2} = [E]$$

Both sides of $\Delta E = hf$ agree.

A quick check. Unit check:

$h$'s unit is $\text{J s}$.

$\text{J} = \text{kg m}^2\text{ s}^{-2}$, so:

$$\text{J s} = \text{kg m}^2\text{ s}^{-1}$$

This matches $ML^2T^{-1}$.

Conclusion

$[h] = ML^2T^{-1}$

The dimension of $h = \dfrac{E}{f}$ is $ML^2T^{-1}$.

Does it make sense? Yes, the only matching option is (5).

Remember $h$'s unit is $\text{J s}$.

A joule's dimension $ML^2T^{-2}$ times $T$ gives $ML^2T^{-1}$.

Angular momentum has this same dimension.

Exam tip