Q 17PE - College Physics (Urone)

Question

Question: (a) Calculate the number of photoelectrons per second ejected from a \(1.00\,{\rm{m}}{{\rm{m}}^{\rm{2}}}\) area of sodium metal by \(500\,{\rm{nm EM}}\) radiation having an intensity of \(1.30\,{\rm{kW/}}{{\rm{m}}^{\rm{2}}}\) (the intensity of sunlight above the Earth’s atmosphere). (b) Given that the binding energy is\(2.28\,{\rm{eV}}\), what power is carried away by the electrons? (c) The electrons carry away less power than brought in by the photons. Where does the other power go? How can it be recovered?

Step-by-Step Solution

Answer

(a) Number of photoelectron \(n = 3.27 \times {10^{15}}\,{\rm{photona}}{{\rm{s}}^{{\rm{ - 1}}}}\).

(b) Power carried by electrons is \({P_{total{\rm{ }}}} = 0.15\,{\rm{mW}}\).

(c) This energy difference is equal to the mass that converts into binding energy of the sodium metal and heat energy. Some energy also used to excite the electrons from lower energy state to higher energy state. When electrons come back from a higher energy state to lower energy state, this energy is released again as photon or heat energy.

Step 1: Wavelength of the ejected electron from sodium metal

As we know that the intensity is defined as the power per unit area and is given by

\(I = \frac{P}{A}\)                                                                   ...(1)

Where P is the power of the radiation and

A is the surface area of the metal.

Therefore from the equation, the power at the surface of the metal is given by

\(P = IA\)                                                                              ...(2)

Now, from the equation, the energy of a photon is given by

\(E = hf\)

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

\(f = \frac{c}{\lambda }\)                                                       ...(3)

Therefore, from the equation, we get

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

The wavelength of the ejected electron from sodium metal is

\(\begin{array}{c}\lambda  = 500\,{\rm{nm}}\\ &= 500 \times {10^{ - 9}}\,{\rm{m}}\end{array}\).

Step 2: Binding energy of the electron in calcium metal

Surface area of the sodium metal is

\(\begin{array}{c}A = 1.00\,{\rm{m}}{{\rm{m}}^{\rm{2}}}\\ &= 1.00 \times {10^{ - 6}}\,{{\rm{m}}^{\rm{2}}}\end{array}\)

Intensity of the EM radiations

\(\begin{array}{c}I = 1.30\,{\rm{kW}}{{\rm{m}}^{{\rm{ - 2}}}}\\ &= 1.30 \times {10^3}\,{\rm{W}}{{\rm{m}}^{{\rm{ - 2}}}}\end{array}\)

Here, binding energy of the electron in calcium metal is

\(BE = 2.20\,{\rm{eV}}\)

Since, we know

\(1.00\;\,{\rm{J}} = 6.242 \times {10^{18}}\,{\rm{eV}}\)

Therefore we get

\(\begin{array}{c}BE = 2.20\,{\rm{eV}} \times \frac{{1.00\,{\rm{J}}}}{{6.242 \times {{10}^{18}}\,{\rm{eV}}}}\\ &= 3.52 \times {10^{ - 19}}\,{\rm{J}}\end{array}\)

Step 3: The number of ejected electrons per second

(a) 

Hence, from equation, the power at the surface of the sodium metal is given by

\(P = IA\)

 

Substitute all the value in the above equation

\(\begin{array}{c}P = 1.30 \times {10^3}\,{\rm{W}}{{\rm{m}}^{{\rm{ - 2}}}} \times 1.00 \times {10^{ - 6}}\,{{\rm{m}}^{\rm{2}}}\\ &= 1.30 \times {10^{ - 3}}\,{\rm{W}}\end{array}\)

 

Now from equation, the energy of the ejected electron from sodium metal surface is

\(E = \frac{{hc}}{\lambda }\)

 

Substitute all the value in the above equation

\(\begin{array}{c}E = \frac{{6.626 \times {{10}^{ - 34}}\,{\rm{Js}} \times 3.00 \times {{10}^8}\,{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}}}{{500 \times {{10}^{ - 9}}\,{\rm{m}}}}\\ &= 3.98 \times {10^{ - 19}}\,{\rm{J}}\end{array}\)

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

\(n = \frac{P}{E}\)

 

Substitute all the value in the above equation

\(\begin{array}{c}n = \frac{{1.30 \times {{10}^{ - 3}}\,{\rm{W}}}}{{3.98 \times {{10}^{ - 19}}\,{\rm{J}}}}\\ = 3.27 \times {10^{15}}\,{\rm{photona}}{{\rm{s}}^{ - 1}}\end{array}\)

Hence, \(n = 3.27 \times {10^{15}}\,{\rm{photona}}{{\rm{s}}^{{\rm{ - 1}}}}\)

Step 4: The power that carried away by the electrons

(b) 

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

\(n = 3.27 \times {10^{15}}\,{\rm{photona}}{{\rm{s}}^{{\rm{ - 1}}}}\)

Hence, the power is carried away to eject n electrons is

\({P_e} = nBE\)

 

Substitute all the value in the above equation

\(\begin{array}{c}{P_e} = 3.27 \times {10^{15}}\,{\rm{photona}}{{\rm{s}}^{ - 1}} \times 3.52 \times {10^{ - 19}}\,{\rm{J}}\\ &= 1.15 \times {10^{ - 3}}\,{\rm{J}}{{\rm{s}}^{{\rm{ - 1}}}}\\ &= 1.15 \times {10^{ - 3}}\,{\rm{W}}\\ &= 1.15\,{\rm{mW}}\end{array}\)

Therefore, the power that carried away by the electrons is

\(\begin{array}{c}{P_{total{\rm{ }}}} = P - {P_e}\\ &= 1.30 \times {10^{ - 3}}\,{\rm{W}} - 1.15 \times {10^{ - 3}}\,{\rm{W}}\\ &= 0.15 \times {10^{ - 3}}\,{\rm{W}}\\ &= 0.15\,{\rm{mW}}\end{array}\)

Hence, \({P_{total{\rm{ }}}} = 0.15\,{\rm{mW}}\)

Step 5: Electrons came back

(c) 

Here, from the calculation in part (b) we can see the electrons carry away less power than brought in by the photons. This energy difference is equal to the mass that converts into binding energy of the sodium metal and heat energy. 

Some energy also used to excite the electrons from lower energy state to higher energy state.

When electrons come back from higher energy state to lower energy state, this energy is released again as a photon or heat energy.