2021 A/L Physics Past Paper · Question 29 (English Medium)

Answer: (1) Only (A) is true.

What the question asks

What we know:

The same metal is lit by three kinds of light, one at a time: blue, red and yellow. Each light has one colour only (monochromatic).

What each letter means:

$f$ ($\text{Hz}$): the frequency of the light.

$f_0$ ($\text{Hz}$): the threshold frequency of the metal, the smallest frequency that can eject electrons.

$\lambda$ ("lambda", $\text{m}$): the wavelength of the light.

$c$ ($\text{m s}^{-1}$): the speed of light, the same for every colour.

$f_R$, $f_Y$, $f_B$ ($\text{Hz}$): the frequencies of red, yellow and blue light.

Find which of the statements (A), (B) and (C) are always true.

The plan: put the three colours in order of frequency. Then test each statement with the rule for the threshold frequency.

Relevant theory

The photoelectric emission takes place only when the frequency of the falling light is greater than a certain minimum value known as the threshold frequency.

The wavelengths from 380 nm (violet) to 760 nm (red) belong to this range colour series red, orange, yellow, green, blue, indigo and violet which can be identified by the eye is called the visible spectrum.

However, in a vacuum or in free space they progress with the same speed as the speed of light

Measurement of the frequency and the wavelength of waves enable their velocity to be calculated.

Reading the options

First, the order of the colours. Every colour travels at the same speed $c$. Speed is frequency times wavelength:

$$c = f\lambda$$

This says: if $c$ stays the same, a shorter wavelength must have a higher frequency.

Red has the longest wavelength, about $760\ \text{nm}$. Yellow is shorter. Blue is shorter still.

So the frequencies go the other way:

$$f_R < f_Y < f_B$$

This says: red has the lowest frequency and blue the highest.

Now the rule. Think of a fair ride that lets in only children taller than a line on the wall. The line is $f_0$.

If a short child gets in, then a taller child surely gets in. But if a tall child gets in, a shorter child may still be too short.

Light is the same. If a lower frequency is above $f_0$, every higher frequency is too. A higher frequency working tells us nothing sure about a lower one.

Reasoning and calculation

If red light ejects photoelectrons, then blue light should also eject photoelectrons.

Red works, so $f_R > f_0$. Blue has an even higher frequency, so $f_B > f_0$ too. Blue must eject electrons.

A

If yellow light ejects photoelectrons, then red light should also eject photoelectrons.

Yellow works, so $f_Y > f_0$. But red has a lower frequency than yellow. It may be below $f_0$. Red need not eject electrons.

B

If blue light ejects photoelectrons, then red light should also eject photoelectrons.

Blue works, so $f_B > f_0$. Red has a much lower frequency. It may be below $f_0$. Red need not eject electrons.

C

Only (A) goes from a lower frequency up to a higher one, so only (A) is sure.

(B) and (C) go down to red, the lowest frequency, so they are not sure. The answer is option (1).

Why the other options fail

Red has a lower frequency than yellow, so (B) is not certain. (A) is true.

(B) goes from yellow down to red, a lower frequency. It is not certain.

(B) and (C) both go down to red. Red may be below the threshold frequency.

A colour below the threshold frequency ejects no electrons, so (B) and (C) are not certain.

Conclusion

Does it make sense?

A very bright red lamp still ejects nothing from a metal whose $f_0$ is above red. Brightness does not help. Only the frequency matters.

So red is the "weakest" colour here, and only a statement that ends at a higher frequency can be sure.

In frequency, red is lowest, then yellow, then blue. Electrons come out only when $f > f_0$.

If red works, blue surely works, so (A) is true. Yellow or blue working does not make red work, so (B) and (C) are false. The answer is (1).

For photoelectric "if this colour works, then that colour works" questions, put the colours in order of frequency first.

Going up in frequency is always safe. Going down is never certain.

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