Question

Light of wavelength \[4000\dot A\] is incident on a sodium surface for which the threshold wavelength of photoelectrons is \[5420\dot A\]. The work function of sodium is:
A. \[0.57\,eV\]
B. \[1.14\,eV\]
C. \[2.29\,eV\]
D. \[4.58\,eV\]

Answer 1

Hint: The work function of a metal is the smallest amount of energy necessary for electron emission from its surface, and the frequency of light corresponding to this minimal energy is called threshold frequency, and the associated wavelength is called threshold wavelength.

Formula Used:
Energy, \[E = h\nu = \dfrac{{hc}}{\lambda }\]
Equation for photoelectric effect, \[h\nu = {\phi _o} + {E_k}\]
Work function, \[{\phi _o} = h{\nu _o} = \dfrac{{hc}}{{{\lambda _o}}}\]
Where,
E = energy of incident radiation
\[{\phi _o}\]= Work function of the metal
\[{E_k}\]= energy of the emitted photoelectron
\[c{\rm{ }}\] = speed of light = \[3 \times {10^8}m/s\]
\[\nu \]= frequency of the light,
\[{\nu _o}\]= threshold frequency,
\[\lambda \]= wavelength of the light,
\[{\lambda _o}\]= threshold wavelength

Complete step by step solution:
Given: In this photoelectric experiment, incident radiation \[\lambda = 4000\dot A\] and the threshold wavelength for sodium, \[{\lambda _o} = 5420\dot A\]. We need to determine the work function of this metal.
Equation for photoelectric effect is,
\[h\nu = {\phi _o} + {E_k}\]---- (1)
\[\Rightarrow {\phi _o} = \dfrac{{hc}}{{{\lambda _o}}}\]----(2)
We know, \[1\dot A = {10^{ - 10}}m\]
\[{\phi _o} = \dfrac{{6.64 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{5420 \times {{10}^{ - 10}}}}\]
\[\Rightarrow {\phi _o} = \dfrac{{6.64 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{5420 \times {{10}^{ - 10}}}}\]
Solving this will give the value of work function in joules. We need to determine the work function in units of eV.
We know that \[1\,eV = 1.6 \times {10^{ - 19}}J\]
So,
\[{\phi _o} = \dfrac{{6.64 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{5420 \times {{10}^{ - 10}} \times 1.6 \times {{10}^{ - 19}}}} \\ \]
\[\therefore {\phi _o} = 2.29\,eV\]

Hence option C is the correct answer.

Additional information: Einstein developed the photoelectric effect in 1905, for which he also received the Nobel Prize in Physics. When sufficient energy light is incident on a metal surface, the photons of the incident light give energy to the electrons on the metallic surface, according to its theory. If this energy exceeds the metal's threshold energy, the electrons become sufficiently energetic to exit the metal surface and be expelled. These electrons are known as photoelectrons since their energy is less than that of the incoming light because some of the energy is used to overcome the barrier energy or the work function.

Note: Work function is a property of the metal that depends on the metal. The remaining energy after the work function is responsible for emission of the electron from the surface. This extra energy is converted to kinetic energy which enables the electron to emit from the metal surface. If we know the energy of the incident radiation and the threshold, we can determine the additional kinetic energy and the stopping potential using \[{E_k} = e{V_o}\].