Question

A laser with a power output of $2.00\mathrm{m}\mathrm{W}$at a wavelength of $400\mathrm{n}\mathrm{m}$ is projected onto calcium metal. (a) How many electrons per second are ejected? (b) What power is carried away by the electrons, given that the binding energy is $2.71\mathrm{e}\mathrm{V}$?

Short answer

(a) Electrons ejected per second $\mathrm{n}=4.02\times 10^{15}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{t}\mathrm{o}\mathrm{n}\mathrm{a}{\mathrm{s}}^{-1}$

(b) Power carried by electrons ${\mathrm{P}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=0.250\mathrm{m}\mathrm{W}$

Step-by-step solution

Number of electrons ejected per second

From equation, the maximum kinetic energy of the ejected electron from any metal surface is

$K{E}_e=hf-BE$ ...(1)

Where$h=6.626\times {10}^{-34}\mathrm{J}\mathrm{s}$is Planck's constant,

$f$is the frequency of the incident photon,

$hf$is the photon's energy, and

The relation between frequency$f$and wavelength$\lambda$is given by

$f=\frac{c}{\lambda }$ ...(2)

From equation, the energy of a photon is given by

$E=hf$ ...(3)

Therefore, from equation, we get

$E=\frac{hc}{\lambda }$ ...(4)

Wavelength of the ejected electron from calcium metal is

$\begin{array}{rl}\lambda & =400\mathrm{n}\mathrm{m}\\ & =400\times {10}^{-9}\mathrm{m}\end{array}$

Mass of the electron is

$m=9.11\times {10}^{-31}\mathrm{k}\mathrm{g}$

The power output of the laser is

$\begin{array}{rl}{P}_{out}& =2.00\mathrm{m}\mathrm{W}\\ & =2.00\times {10}^{-3}\mathrm{W}\end{array}$

Here, binding energy of the electron in calcium metal is

$BE=2.71\mathrm{e}\mathrm{V}$

Since, we know

$1.00\mathrm{J}=6.242\times {10}^{18}\mathrm{e}\mathrm{V}$

Calculate the number of electrons ejected per second

Therefore we get

$\begin{array}{rl}BE& =2.71\mathrm{e}\mathrm{V}\times \frac{1.00\mathrm{J}}{6.242\times {10}^{18}\mathrm{e}\mathrm{V}}\\ & =4.34\times {10}^{-19}\mathrm{J}\end{array}$

Now from equation, the energy of the ejected electron from calcium is

$E=\frac{hc}{\lambda }$

Substitute all the value in the above equation

$\begin{array}{rl}E& =\frac{6.626\times {10}^{-34}\mathrm{J}\mathrm{s}\times 3.00\times {10}^8\mathrm{m}{\mathrm{s}}^{-1}}{400\times {10}^{-9}\mathrm{m}}\\ & =4.97\times {10}^{-19}\mathrm{J}\end{array}$

Now we know the power is defined as the energy per second. Therefore, the number of ejected electrons per second is given by

$n=\frac{P_{out}}{E}$

Substitute all the value in the above equation

$\begin{array}{rl}n& =\frac{2.00\times {10}^{-3}\mathrm{W}}{4.97\times {10}^{-19}\mathrm{J}}\\ & =4.02\times {10}^{15}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{t}\mathrm{o}\mathrm{n}\mathrm{a}{\mathrm{s}}^{-1}\end{array}$

Therefore, the number of ejected electrons per second is given by $n=4.02\times {10}^{15}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{t}\mathrm{o}\mathrm{n}\mathrm{a}{\mathrm{s}}^{-1}$

Power is carried away by electrons

(b)

Now from equations, the kinetic energy of the ejected electron is

$K{E}_e=\frac{hc}{\lambda }-BE$

Substitute all the value in the above equation

$\begin{array}{rl}K{E}_e& =\frac{6.626\times {10}^{-34}\mathrm{J}\mathrm{s}\times 3.00\times {10}^8\mathrm{m}{\mathrm{s}}^{-1}}{400\times {10}^{-9}\mathrm{m}}-4.34\times {10}^{-19}\mathrm{J}\\ & =6.30\times {10}^{-20}\mathrm{J}\end{array}$

Now as we know the power is defined as the energy per unit time and is given by

$P=\frac{KE}{t}$

The number of ejected electrons per second is

$n=4.02\times {10}^{15}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{t}\mathrm{o}\mathrm{n}\mathrm{a}{\mathrm{s}}^{-1}$

The power is carried away by n electrons is

$\begin{array}{rl}{P}_{total}& =n\times KE\\ & =4.02\times {10}^{15}\mathrm{p}\mathrm{h}\mathrm{o}\mathrm{t}\mathrm{o}\mathrm{n}\mathrm{a}{\mathrm{s}}^{-1}\\ & =2.50\times {10}^{-1}\mathrm{J}{\mathrm{s}}^{-1}\\ & =2.50\times {10}^{-4}\mathrm{W}\\ & =0.250\mathrm{m}\mathrm{W}\end{array}$

Hence, the power is carried away by n electrons is $P_{total}=0.250\mathrm{m}\mathrm{W}$