When the light of frequency 2 \[{{v}_{0}}\] (where \[{{v}_{0}}\] is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is \[{{v}_{1}}\]. When the frequency of the incident radiation is increased to 5\[{{v}_{0}}\], the maximum velocity of electrons emitted from the same plate is \[{{v}_{2}}\]. The ratio of $v_1$ to $v_2$?
A. 4:1
B. 1:2
C. 2:1
D. 1:4

Hint: Here we are given that a light is incident on the metal. In accordance with photo electric effect, if the frequency of the incident radiation is greater than the threshold frequency of the metal, then photo electrons will get emitted. This was given by Albert Einstein.

Complete step by step answer:
We can use the equation of photoelectric effect to solve this problem. From photoelectric effect, Einstein equation is given by \[h\nu =h{{\nu }_{0}}+KE\]
Here KE is the kinetic energy given by \[\dfrac{m{{v}^{2}}}{2}\]. So, putting It in above equation we get, \[h\nu =h{{\nu }_{0}}+\dfrac{m{{v}^{2}}}{2}\]
Rewriting it, \[h\nu -h{{\nu }_{0}}=\dfrac{m{{v}^{2}}}{2}\]
Putting the values,
\[\dfrac{m{{v}_{1}}^{2}}{2}=2h{{\nu }_{0}}-h{{\nu }_{0}}\]
\[\dfrac{m{{v}_{1}}^{2}}{2}=h{{\nu }_{0}}\]----(1)
For second case, \[\dfrac{m{{v}_{2}}^{2}}{2}=5h{{\nu }_{0}}-h{{\nu }_{0}}\]
\[\dfrac{m{{v}_{2}}^{2}}{2}=4h{{\nu }_{0}}\]---(2)
Dividing eq (1) by (2)
\[\dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{1}{2}\]

So, the correct answer is “Option B”.

Additional Information:
Photoelectric effect is the phenomenon of emission of electrons when radiation of a given frequency is incident on the metal. There exists a frequency called threshold frequency below which no photo electric effect can take place. Threshold frequency changes with the change in the material.

Note:
Below threshold frequency, the photoelectric effect does not happen. Also, upon increasing the intensity of the radiation, the kinetic energy of the photon emitted electrons does not change. Threshold frequency changes with the change in the material.