"Dual Nature of Matter and Radiation" — Practice

Question 1

A metal is illuminated with light above its threshold frequency. What happens to the stopping potential and to the saturation current if the intensity is doubled?

Accepted Answer

The stopping potential is unchanged, because it depends only on photon energy and hence only on frequency. The saturation current doubles, because twice as many photons arrive per second and each can free one electron.

Question 2

Why is there no time lag in photoelectric emission even with very dim light?

Accepted Answer

Because energy is not accumulated. A single photon either carries enough energy to free an electron in one all-or-nothing exchange or it does not. Dim light means fewer photons per second, not weaker photons, so emission begins within a nanosecond of the first adequate photon arriving.

Question 3

Is the threshold a maximum or a minimum wavelength, and why does the distinction matter?

Accepted Answer

It is a maximum wavelength, corresponding to a minimum frequency. Light of longer wavelength has lower frequency and therefore lower photon energy, so it ejects nothing at any intensity. Reversing this is a common sign error.

Question 4

A metal of work function 2.0 eV is illuminated with 400 nm light. Find the maximum kinetic energy, the stopping potential and the threshold wavelength.

Accepted Answer

Using $hc = 1240$ eV nm, the photon energy is 3.1 eV, so $K_{max} = 1.1$ eV and $V_0 = 1.1$ V. The threshold wavelength is $1240/2.0 = 620$ nm, above which nothing is emitted.

Question 5

A 5 mW laser emits at 632.8 nm. Find the number of photons emitted per second.

Accepted Answer

Each photon carries $hc/\lambda \approx 3.14	imes10^{-19}$ J. Dividing the power by that gives about $1.6	imes10^{16}$ photons per second, which is why the discreteness of light is entirely undetectable by eye.

Question 6

An electron is accelerated through 100 V. Find its de Broglie wavelength, and compare it with a proton accelerated through the same potential.

Accepted Answer

$\lambda = 12.27/\sqrt{100} = 1.227$ Å for the electron. Since the charges are equal, $\lambda \propto 1/\sqrt{m}$, and a proton is 1836 times heavier, so its wavelength is smaller by about 42.8, giving 0.0287 Å.

Question 7

An X-ray tube runs at 50 kV. Find the shortest wavelength produced and state the effect of changing the target metal.

Accepted Answer

The cut-off is $1240/50000$ nm $= 0.0248$ nm, or 0.248 Å. Changing the target has no effect on the cut-off, which is set entirely by the accelerating voltage since that fixes the largest energy any single electron can deliver. The target only determines the characteristic lines superimposed on the continuous spectrum.

Question 8

Light of 300 nm and 400 nm give stopping potentials of 2.13 V and 1.10 V. Find the work function and Planck's constant.

Accepted Answer

The photon energies are 4.13 eV and 3.10 eV, so subtracting the two photoelectric equations eliminates the work function and gives a consistent difference of 1.03 eV. The work function is $4.13 - 2.13 = 2.00$ eV. Using the frequency difference of $0.25	imes10^{15}$ Hz, $h \approx 6.6	imes10^{-34}$ J s.

Question 9

A photon and an electron both have a wavelength of 1 nm. Compare their momenta and energies, and explain the difference.

Accepted Answer

Their momenta are equal, since both are $h/\lambda$. The photon's energy is $hc/\lambda = 1240$ eV, while the electron's kinetic energy is $p^{2}/2m \approx 1.5$ eV. The photon carries roughly 800 times more energy despite identical momentum, because a massless photon obeys $E = pc$ while a slow massive particle obeys $E = p^{2}/2m$.

Question 10

What does the double slit experiment performed with single electrons show, and why can neither classical model describe it alone?

Accepted Answer

Each electron arrives as a single localised dot on the screen, which is particle behaviour, yet after many thousands the dots build into an interference pattern, which is wave behaviour. Since only one electron is in the apparatus at a time, each must interfere with itself. Attempting to detect which slit it passed through destroys the pattern however gently the measurement is made, so the two descriptions are mutually exclusive rather than merely alternative.

Practice — "Dual Nature of Matter and Radiation"

Sign in to save attempts to your dashboard and track your accuracy over time

Find a verified tutor

Join a live course

Browse coaching centres

About

Categories

Support

Documentation

Get Study Tips & Updates

Download Our App

Download for iOS

Download for Android

Popular Cities