Question

The work function of a metal is \[1.6 \times {10^{ - 19}}J\]. When the metal surface is illuminated by the light of wavelength \[{\rm{6400}}\mathop {\rm{A}}\limits^{\rm{0}} \] , then find the maximum kinetic energy of emitted photoelectrons. Given that \[\left( {h = 6.4 \times {{10}^{ - 34}}Js} \right)\]
A. \[14 \times {10^{ - 19}}J\]
B. \[2.8 \times {10^{ - 19}}J\]
C. \[1.4 \times {10^{ - 19}}J\]
D. \[1.4 \times {10^{ - 19}}eV\]

Answer 1

Hint: When light is incident on a metal surface, electrons are emitted known as photoelectrons. This phenomenon is called the photoelectric effect. The minimum energy under which the emission of photoelectrons takes place is known as the threshold energy. Work function is the minimum energy required to remove one electron from a metal surface. Work function can also be said as the characteristics of the surface and not the complete metal. As we know that the energy of the incident light or photons should be more than the work function of the metal for the photoemission to take place.

Formula Used:
To find the kinetic energy of the emitted electron we have,
\[K.{E_{\max }} = E - W\]
Where, W is the work function and E is the energy of a photon.

Complete step by step solution:
As we know that the kinetic energy of the emitted electron is,
\[K.{E_{\max }} = E - W\]
\[\Rightarrow K.{E_{\max }} = h\nu - W\]
\[\Rightarrow K.{E_{\max }} = \dfrac{{hc}}{\lambda } - W\]
Substitute the values of Planck’s constant \[\left( {h = 6.4 \times {{10}^{ - 34}}Js} \right)\], speed of light, wavelength, \[\lambda {\rm{ = 6400}}\mathop {\rm{A}}\limits^{\rm{0}} \] and work function, \[W = 1.6 \times {10^{ - 19}}J\] in the above equation.

After substitution we get,
\[K.{E_{\max }} = \dfrac{{6.4 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{6400 \times {{10}^{ - 10}}}} - 1.6 \times {10^{ - 19}}\]
\[\Rightarrow K.{E_{\max }} = \dfrac{{6.4 \times 3 \times {{10}^{ - 26}}}}{{6.4 \times {{10}^{ - 7}}}} - 1.6 \times {10^{ - 19}}\]
\[\Rightarrow K.{E_{\max }} = 3 \times {10^{ - 19}} - 1.6 \times {10^{ - 19}}\]
\[\therefore K.{E_{\max }} = 1.4 \times {10^{ - 19}}J\]
Therefore, the maximum kinetic energy of emitted photoelectrons is \[1.4 \times {10^{ - 19}}J\].

Hence, Option C is the correct answer.

Note:Remember that, if the energy is given in eV, then we need to convert the given value of energy from eV to joules by multiplying the charge of an electron while solving this problem. Moreover, metals contain many free electrons that are loosely bound to the atom. The typical value of work function varies from 2ev to 6eV.