Question

In a photoelectric effect experiment, irradiation of a metal with light of frequency $5.2 \times 10^{14}sec^{-1}$ yields electrons with maximum kinetic energy $1.3 \times 10^{-19}J$. calculate the $v_0$ of the metal.
A) $3.2 \times 10^{14}se{c^{-1}}$
B) $1.6 \times 10^{14}se{c^{-1}}$
C) $6.4 \times 10^{14}se{c^{-1}}$
D) $3.2 \times 10^{15}se{c^{-1}}$

Answer 1

Hint: We know that the photoelectric effect states that when a beam of light of certain minimum frequency is allowed to strike the metal surface in a vacuum then there is ejection of electrons from the metal surface.

Formula used:
$K.E = h\left( {v - {v_0}} \right)$

Complete answer:
If the frequency of light is more than the threshold frequency then some of the energy gets consumed to eject the electron from an atom and extra energy render the electron with kinetic energy i.e $\dfrac{1}{2}m{v^2}$.
i.e Total energy (E) = I+K.E
where I is the ionization energy required to release the electron and K.E is the kinetic energy of the electron. We can write it as.
$hv = h{v_0} + \dfrac{1}{2}m{v^2}$
Where, h is the planck's constant
${v_0}$ is the threshold frequency
$v$ is the frequency
First we have to write given values.
Frequency of metal= $5.2 \times {10^{14}}{s^{ - 1}}$
Kinetic energy = $1.3 \times {10^{ - 19}}$
$ \Rightarrow h{v_0} = hv - K.E$
Divided the h above equation we get.
$ \Rightarrow {v_0} = v - \dfrac{{K.E}}{h}$
Also we know that plan`s constant is
h= 6.626×10-34
substituting above values we get.
$ \Rightarrow {v_0} = 5.2 \times {10^{14}} - \dfrac{{1.3 \times {{10}^{ - 19}}}}{{6.626 \times {{10}^{ - 34}}}}$
 Solving mathematically
$\therefore {v_0} = 3.24 \times {10^4}{s^{ - 1}}$
Hence, the threshold frequency for the metal is $3.24 \times {10^4}{s^{ - 1}}$.

The given option A is the correct answer.

Additional information:
Threshold frequency states that the minimum frequency of radiation that will produce photoelectric effect.
Kinetic energy is nothing but is the energy an object has because of its motion. It depends on mass and speed achieved.
There are two forms of energy : potential energy and kinetic energy. Potential energy refers to stored energy and kinetic energy is energy in motion.

Note:
Increase in the intensity of incident radiation increases the rate of emission not the energy of the photoelectron. In simple words the number of photoelectrons ejected is proportional to the intensity of incident radiation. The alkali metals having lower ionization energy readily show the photoelectric effect and caesium having lowest ionization energy is the best way to show a photo electric effect.