MediumChemistryMiddle Stage (6–8)Structure of the Atom › Subatomic Particles and Atomic ModelsNumericalJEE Main

Light of wavelength 330 nm falls on a clean metal surface.

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Light of wavelength $330\ \text{nm}$ falls on a clean metal surface. The work function of the metal is $2.25\ \text{eV}$ , and the beam carries a power of $2.0\ \text{mW}$ . Take $h = 6.626 \times 10^{-34}\ \text{J s}$ , $c = 3.0 \times 10^{8}\ \text{m s}^{-1}$ , $m_e = 9.109 \times 10^{-31}\ \text{kg}$ and $1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}$ .

What is the energy of one photon of this light, in joules and in electronvolts?
What is the largest kinetic energy of the electrons that leave the surface?
What is the largest speed of these electrons?
Up to what wavelength does this metal give out electrons at all (its threshold wavelength)?
How many photons strike the surface each second?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Find the energy of one photon, take away the work function, and use the power for the count.

(a) One photon.

\displaystyle E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{330 \times 10^{-9}}\ \text{J} = 6.024 \times 10^{-19}\ \text{J}

In electronvolts, $E = \dfrac{6.024 \times 10^{-19}}{1.602 \times 10^{-19}}\ \text{eV} = 3.76\ \text{eV}$ .

(b) The fastest electrons. $K_{\max} = E - W_0 = 3.76\ \text{eV} - 2.25\ \text{eV} = 1.51\ \text{eV}$ , that is $2.419 \times 10^{-19}\ \text{J}$ .

(c) Their speed. From $\tfrac12 m_e v^2 = K_{\max}$ :

\displaystyle v_{\max} = \sqrt{\frac{2K_{\max}}{m_e}} = \sqrt{\frac{2 \times 2.419 \times 10^{-19}}{9.109 \times 10^{-31}}}\ \text{m s}^{-1} = 7.29 \times 10^{5}\ \text{m s}^{-1}

(d) Threshold. $W_0 = 2.25 \times 1.602 \times 10^{-19}\ \text{J} = 3.605 \times 10^{-19}\ \text{J}$ , so

\displaystyle \lambda_0 = \frac{hc}{W_0} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{3.605 \times 10^{-19}}\ \text{m} = 551.5\ \text{nm}

Light of longer wavelength gives no electrons, however bright it is.

(e) Photons each second.

\displaystyle N = \frac{P}{E} = \frac{2.0 \times 10^{-3}\ \text{J s}^{-1}}{6.024 \times 10^{-19}\ \text{J}} = 3.32 \times 10^{15}\ \text{photons per second}

Check. By frequencies: $\nu = c/\lambda = 9.091 \times 10^{14}\ \text{Hz}$ and $\nu_0 = W_0/h = 5.440 \times 10^{14}\ \text{Hz}$ , and $h(\nu - \nu_0)$ gives the same $2.419 \times 10^{-19}\ \text{J}$ .

Watch out. Do not use electronvolts in $\tfrac12 m_e v^2$ : turn the energy into joules first.

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← For N₂O₄(g) ⇌ 2NO₂(g), Δ H° = +57.2 kJ/mol and Δ S° = +175.8 J/(K mol).A hydrogen-like ion has one electron and a nucleus of charge +Ze. →

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