Question

The photoelectric threshold wavelength of silver is $3250 \times {10^{ - 10}}m$ . The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times {10^{ - 10}}m$ , is
[Given $h = 4.14 \times {10^{ - 15}}eVs$ and $c = 3 \times {10^8}m{s^{ - 1}}$ ]
\[
  A.{\text{ }}6 \times {10^5}m{s^{ - 1}} \\
  B.{\text{ }}0.6 \times {10^6}m{s^{ - 1}} \\
  C.{\text{ }}61 \times {10^3}m{s^{ - 1}} \\
  D.{\text{ }}0.3 \times {10^6}m{s^{ - 1}} \\
 \]

Answer 1

Hint: In order to find the required velocity of given question, here we will use the formula of kinetic energy in terms of threshold wavelength, so first we will convert the formula of kinetic energy in terms of threshold wavelength then by substituting the values we will proceed further.
Formula used- \[{K_{\max }} = h\upsilon - {\phi _0},\upsilon = \dfrac{c}{\lambda },{\phi _0} = h{\upsilon _0},K = \dfrac{1}{2}m{v^2}\]

Complete Step-by-Step solution:
Given that:
Threshold wavelengths $3250 \times {10^{ - 10}}m$
Here for the problem we have:
\[
  h = 4.14 \times {10^{ - 15}}eVs \\
  c = 3 \times {10^8}m{s^{ - 1}} \\
  {\lambda _0} = 3250 \times {10^{ - 10}}m = 3250{\text{A}} \\
  \lambda = 2536 \times {10^{ - 10}}m = 2536{\text{A}} \\
  m = 9.1 \times {10^{ - 31}}Kg \\
  hc = 4.14 \times {10^{ - 15}}eVs \times 3 \times {10^8}m{s^{ - 1}} \\
  hc = 12420eV{\text{A}} \\
 \]
We know that
The maximum kinetic energy is given as
$
  {K_{\max }} = h\upsilon - {\phi _0} \\
  {K_{\max }} = h\upsilon - h{\upsilon _0} \\
  {K_{\max }} = \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}}..........(1) \\
 $
Here ${\lambda _0}$ is the threshold wavelength
As we know that the kinetic energy of any object is given by the formula:
$K = \dfrac{1}{2}m{v^2}$
Let use substitute the formula of kinetic energy in equation (1)
$
  \because {K_{\max }} = \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}} \\
   \Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}} \\
 $
As we have all the value given, so let us substitute the values in the question in order to find the velocity
$
  \because \dfrac{1}{2}m{v^2} = \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}} \\
   \Rightarrow \dfrac{1}{2}m{v^2} = hc\left[ {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _0}}}} \right] \\
   \Rightarrow {v^2} = \dfrac{{2hc}}{m}\left[ {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _0}}}} \right] \\
   \Rightarrow {v^2} = \dfrac{{2 \times 12420}}{m}\left[ {\dfrac{1}{{2536}} - \dfrac{1}{{3250}}} \right]eV \\
   \Rightarrow {v^2} = \dfrac{{2.152eV}}{m} \\
   \Rightarrow {v^2} = \dfrac{{2.152 \times 1.6 \times {{10}^{ - 19}}}}{{9.1 \times {{10}^{ - 31}}}} \\
   \Rightarrow v \approx 6 \times {10^5}m{s^{ - 1}} \\
   \Rightarrow v \approx 0.6 \times {10^6}m{s^{ - 1}} \\
 $
Hence, the velocity of the electron ejected from a silver surface is $ \approx 0.6 \times {10^6}m{s^{ - 1}}$ ,
So, the correct answer is options A and B, because both are the same.

Note- The photoelectric effect is the emission of electrons when electromagnetic radiation, such as light, hits a material. Electrons emitted in this manner are called photoelectrons. This phenomenon is commonly studied in electronic physics. An object's kinetic energy is the energy it retains due to its motion. It is defined as the work necessary to speed up a body of a given mass from rest to its specified speed.