Question

Metal	Li	Na	K	Mg	Cu	Ag	Fe	Pt	W
W(eV)	2	4.2	3.2	2.3	7.4	8.4	3.4	7.63	4.75

 The energy required to overcome the attraction forces on the electrons, w of some metals is listed below. The number of metals showing photoelectric effect when the light of 300nm wavelength falls on it is(\[1eV = 1.6 \times 1{0^{ - 19}}\,J\]).
A) 8
B) 5
C) 4
D) 6

Answer 1

Hint: The electrons from the metal surface are emitted by the application of light energy this is known as the photoelectric effect. The metal is irradiated with light having a frequency greater than the cut of frequency then and then only metal shows the photoelectric effect. Thus, the work function of the metals plays an important role in the photoelectric effect.

Formula used: The photoelectric effect is represented using the equation as follows:
\[h\nu = w + \frac{1}{2}m{v^2}\] (i)
Here, the energy of the light irradiated that is photon is \[h\nu \], the work function of the metal is w, and the kinetic energy of the emitted electron is \[\frac{1}{2}m{v^2}\].
The photon energy is also given as follows:
\[h\nu = h\frac{c}{\lambda }\] (ii)
Here, h is the Plank constant, c is the velocity of light,\[\lambda \] is the wavelength of photon or light irradiated.

Complete step-by-step answer:
Here, the wavelength of light is 300nm and falls on metals. We have to first determine the energy of the photons using the 2nd formula.
\[h\nu = h\frac{c}{\lambda }\]
Here, substitute \[6.626 \times 1{0^{ - 34}}\,Js\] for h, \[3.0 \times 1{0^8}\,m{s^{ - 1}}\]for c, and \[300 \times 1{0^{ - 9}}\,m\] for \[\lambda \].

\[{{h}}\frac{{{c}}}{{{\lambda }}}{{ = 6}}{{.626 \times 1}}{{{0}}^{{{ - 34 }}}}{{J s}} \times \frac{{{{3}}{{.0 \times 1}}{{{0}}^{{8}}}{{ m }}{{{s}}^{{{ - 1}}}}}}{{{{300 \times 1}}{{{0}}^{{{ - 9}}}}{{ m}}}}\]
\[h\frac{c}{\lambda } = 6.626 \times 1{0^{ - 19}}J\,\]
Here, the energy of the photon obtained is . Now, convert photon energy into the eV as follows:
\[1eV = 1.6 \times 1{0^{ - 19}}\,J\]
\[6.626 \times 1{0^{ - 19}}J\,\, = 6.626 \times 1{0^{ - 19}}J\,\, \times \frac{{1 eV}}{{1.6 \times 1{0^{ - 19}}\,J}}\]
\[6.626 \times 1{0^{ - 19}}J\, = 4.14\,eV\]
Thus, the photon energy in electron volt is \[4.14\,eV\].
To show the photoelectric effect photon energy should be greater than the work function. In the question, the work functions of the different metals are given.

Here, Li(lithium), K(potassium), magnesium (Mg), and iron(Fe) have work function less than the energy of the photon. Thus, these 4 metals show the photoelectric effect.

Thus, option(C) is the correct answer for the given question.

Note: The important condition to show the photoelectric effect is the photon energy should be greater than the work function of the metal. The cut-off frequency is the minimum amount of frequency required for the metal to emit the electron from the metal surface.
To convert length from nanometer to meter following conversion factor is used:
\[{{1}}\,{{nm = 1}}{{{0}}^{{{ - 9}}}}\,{{m}}\].