Question

According to Einstein’s photoelectric equation, the plot of the kinetic energy of the emitted photoelectrons from a metal vs. the frequency, of the incident radiation gives a straight line whose slope:
(A) Depends on the nature of the metal used
(B) Depends on the intensity of the radiation
(C) Depends both on the intensity of the radiation and the metal used.
(D) Is the same for all metals and independent of the intensity of the radiation

Answer 1

Hint: The kinetic energy, vs frequency equation can be compared to the equation of a straight line. Intensity of radiation is independent of the frequency and vice versa.
Formula used: In this solution we will be using the following formulae;
\[KE = hf - W\] where \[KE\] is the kinetic energy of the electrons ejected, \[f\] is the frequency of the radiation, \[h\] is the Planck’s constant and \[W\] is called the work function of the metal.

Complete Step-by-Step Solution:
It was observed, in the late 19th century, that when light or any radiation is shone on certain metals, electrons are emitted from the metal. However, the behaviour of these phenomenons at the time did not occur according to the predictions of the theory at the time. Einstein came forth and gave a correct theory which can be summarized according to the equation
\[KE = hf - W\]where \[KE\] is the kinetic energy of the electrons ejected, \[f\] is the frequency of the radiation, \[h\] is the Planck’s constant and \[W\] is called the work function of the metal. The work function is a constant of the metal used.
Hence, if we compare this equation to that of the equation of a straight line given as
\[y = mx + c\]
We can see that the slope \[m\] is equal to \[h\] in the Einstein equation.
Hence, the slope is independent of all intensity or anything but a constant for all metals.

Hence, the correct option is C

Note: Alternatively, without the use of the equation. With the knowledge that the frequency is proportion to the kinetic energy, and that the frequency of radiation has no hidden relationship with the metal nor with the intensity of the radiation, we can conclude, at least, that the kinetic energy is independent of intensity and nature of metal used.