Question

If the work function for a certain metal is \[3.2 \times {10^{ - 19}}J\]and it is illuminated with light of frequency \[8 \times {10^{14}}Hz\].The maximum kinetic energy of the photoelectrons would be: (\[h{\rm{ }}\; = \;6.63 \times {10^{ - 34}}Js\])
A. \[2.1 \times {10^{ - 19}}J\]
B. \[8.5 \times {10^{ - 19}}J\]
C. \[5.3 \times {10^{ - 19}}J\]
D. \[3.2 \times {10^{ - 19}}J\]

Answer 1

Hint: The work function of the metal determines the least energy needed for electron emission from its surface. The frequency of light that corresponds to this minimal energy is known as the threshold frequency, and the associated wavelength is known as the threshold wavelength.

Formula Used:
Energy, \[E = h\nu = \dfrac{{hc}}{\lambda }\]
Equation for photoelectric effect, \[h\nu = {\phi _o} + {E_k}\]
Work function, \[{\phi _o} = h{\nu _o} = \dfrac{{hc}}{{{\lambda _o}}}\]
Maximum kinetic energy, \[{E_k} = e{V_o}\]
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
\[e = 1.6 \times {10^{ - 19}}C\]= electronic charge
\[{V_o}\]= stopping potential

Complete step by step solution:
Given: In this photoelectric experiment, the work function for a certain metal is \[{\phi _o} = 3.2 \times {10^{ - 19}}J\] and frequency of incident radiation, \[\nu = 8 \times {10^{14}}Hz\]. We need to determine the maximum kinetic energy of this metal.
Equation for photoelectric effect is
\[h\nu = {\phi _o} + {E_k}\]---- (1)
\[\Rightarrow {\phi _o} = \dfrac{{hc}}{{{\lambda _o}}}\]----(2)
From equation (1),
\[{E_k} = h\nu - {\phi _o}\]---(3)
\[\Rightarrow {E_k} = (6.63 \times {10^{ - 34}})(8 \times {10^{14}}) - 3.2 \times {10^{ - 19}}\]
Solving this will give the value of maximum kinetic energy in joules.
\[\therefore {E_k} = 2.1 \times {10^{ - 19}}J\]

Hence option A is the correct answer.

Additional information: Einstein discovered the photoelectric effect in 1905, for which he also received the physics Nobel prize. Its theory states that when a light with appropriate energy strikes a metal surface, the photons of the incoming light provide the metallic surface's electrons energy. The electrons become sufficiently energetic to break out from the metal surface and emit when this energy exceeds the threshold energy of the metal. Because some of the energy is used to overcome the barrier energy or the work function, these electrons are known as photoelectrons and have less energy than the energy of the incident light.

Note: The metal's work function is a characteristic that is metal-dependent. The electron is emitted from the surface by the leftover energy following the work function. In order for the electron to emit from the metal surface, the excess energy is transformed to kinetic energy. Using \[{E_k} = e{V_o}\], we can calculate the extra kinetic energy and stopping potential if we know the energy of the incoming radiation and the threshold.