Question

Light of frequency \[8 \times {10^5}{\rm{ Hz}}\] is incident on a substance of photoelectric work function 6.125 eV. The maximum kinetic energy of the emitted photoelectrons is
A. 17 eV
B. 22 eV
C. 27 eV
D. 37 eV

Answer 1

Hint: According to the Einstein equation, the maximum kinetic energy of electrons is equal to the energy of the incident light energy packet minus the work function. Photoelectric is the phenomenon where electrons are ejected from a metal surface when light of sufficient frequency is incident on it. Photoelectrons thus are ejected from a material when light is incident on the surface. Different wavelengths will result in different maximum kinetic energy.

Formula used The energy of the photon is given by the equation is given as:
\[E = h\upsilon = \dfrac{{hc}}{\lambda }\].
Where \[h\] is the Plank’s constant, \[\upsilon \] is the frequency of incident light, c is the speed of light and \[\lambda \] is the wavelength.
The maximum kinetic energy of photoelectrons is given as:
\[K{E_{\max }} = E - \phi \]
Where E is the energy and \[\phi \] is the work function.

Complete step by step solution:
Given Frequency of light, \[\upsilon = 8 \times {10^{15}}Hz\]
Work function, \[\phi \]= 6.125 eV
As we know that the energy of the photon is,
\[E = h\upsilon \]
\[\Rightarrow E = 6.6 \times {10^{ - 34}} \times 8 \times {10^{15}}\]
\[\Rightarrow E= 5.28 \times {10^{ - 18}}J\]
\[\Rightarrow E= 33\,eV\]
By Einstein equation,
\[K{E_{\max }} = E - \phi \]
\[\Rightarrow K{E_{\max }}= 33 - 6.125\]
\[\therefore K{E_{\max }} = 27\,eV\]
Therefore, the maximum kinetic energy of the emitted photoelectrons is \[27\,eV\].

Hence option C is the correct answer.

Note: Always remember that maximum kinetic energy of the ejected electrons depends only on the energy of the incident radiation and independent of the intensity of it. Most of the students make this mistake while solving the problem or applying the concept of photoelectric effect.