Question

For silver metal threshold frequency
${\nu _0}$ is $1.13 \times {10^{17}}Hz$. What is the maximum kinetic energy of photoelectrons produced by shining U.V light of 15.0$\mathop A\limits^ \circ$ wavelength on the metal?
A. $2 \times {10^{17}}J$
B. $5.8 \times {10^{ - 17}}J$
C. $2.9 \times {10^{ - 17}}J$
D. $9 \times {10^{ - 17}}J$

Answer 1

Since the energy possessed by a photoelectron is the sum of threshold energy and kinetic energy, the kinetic energy will be the difference between the total energy and the threshold energy. The threshold energy is the product of the Planck’s constant and the threshold frequency. We can find the frequency of the irradiated light by dividing the speed of light with the wavelength given.

Formula used: $E = {E_0} + K.{E_{\max }}$
Where $E$ is the total energy supplied, ${E_0}$ is the threshold energy and $K.{E_{\max }}$ is the maximum kinetic energy.
$E = h\nu$
Where $h$ is the Planck’s constant and $\nu$ is the frequency of the irradiated light.
$\nu = \dfrac{c}{\lambda }$
Where $c$ is the speed of light and $\lambda$ is the wavelength.

Complete step by step answer:
For any metal, according to the photoelectric effect,
$E = {E_0} + K.{E_{\max }}$
Where $E$ is the total energy supplied, ${E_0}$ is the threshold energy and $K.{E_{\max }}$ is the maximum kinetic energy.
As we know, energy can be expressed as:
$E = h\nu$
Where $h$ is the Planck’s constant and $\nu$ is the frequency of the irradiated light.
Thus, the equation for energy becomes:
$h\nu = h{\nu _0} + K.{E_{\max }}$
$\Rightarrow h\nu - h{\nu _0} = K.{E_{\max }}$
Thus, we get the following equation:
$K.{E_{\max }} = h(\nu - {\nu _0})$
As we know, the frequency of light is related to its wavelength by the equation:
$\nu = \dfrac{c}{\lambda }$
Where $c$ is the speed of light and $\lambda$ is the wavelength.
We know, $c = 3 \times {10^8}m/s$ and it is given that $\lambda$ = 15.0$\mathop A\limits^ \circ$ $= 15 \times {10^{ - 10}}m$. Substituting these, we get:
$\nu = \dfrac{{3 \times {{10}^8}}}{{15 \times {{10}^{ - 10}}}} = 2 \times {10^{17}}Hz$
Substituting this in our equation for kinetic energy, and giving $h = 6.626 \times {10^{ - 34}}Js$ and ${\nu _0} = 1.13 \times {10^{17}}Hz$, we get:
$K.{E_{\max }} = 6.626 \times {10^{ - 34}}(2 \times {10^{17}} - 1.13 \times {10^{17}})$
On solving this, we get:
$K.{E_{\max }} = 6.626 \times {10^{ - 17}} \times 0.87$
$\Rightarrow K.{E_{\max }} = 5.764 \times {10^{ - 17}}J$

Since this answer is approximately equal to $5.8 \times {10^{ - 17}}J$, the correct option to be marked is B.

Note: The photoelectric effect is the phenomenon in which highly energetic photoelectrons are ejected out when light of certain frequency is irradiated on the surface of a few metals. This was first identified by Albert Einstein. Note that at frequencies below the threshold frequency, the metal will not show photoelectric effect, since the energy supplied will not be enough to break away from the pull of the nucleus.