Question

If the frequency of light incident on metal surface is doubled, then the kinetic energy of emitted electrons will be
A. Doubled.
B. Less than double.
C. More than double.
D. Nothing can be used.

Answer 1

Hint: For solving this question we must know photoelectric effect and what is its significance. Then we will move on to calculate the kinetic energy of both the conditions by using the photoelectric equation by Einstein which is given as the energy of the photon will be the sum total of energy needed to remove the electron and the kinetic energy of the emitted electron. Then we will compare the kinetic energy in both cases.
Formula used:
 \[K.E.=h\upsilon -\phi \]

Complete answer:
We should know that, after the concept of photon came out, energy of a photon could be calculated by using planck's equation which is given as,
\[{{E}_{photon}}=h\upsilon \]
Where, \[{{E}_{photon}}\]is the energy of photons in joules.
\[h\] is the planck's constant (\[6.626\times {{10}^{-34}}Js\])
\[\upsilon \] is the frequency of incident light.
Now, let us assume that the initial kinetic energy of the photon be \[K.E\] and the kinetic energy after increasing the frequency be \[K.E'\]. Also, let \[\upsilon \] be the initial frequency and \[2\upsilon \] be the final frequency.
Then, by Einstein’s photoelectric equation,
\[K.E.=h\upsilon -\phi \]
Where, \[K.E\] is the kinetic energy of photons in joules.
\[h\] is the planck's constant (\[6.626\times {{10}^{-34}}Js\])
\[\upsilon \] is the frequency of incident light.
And,\[\phi \]is the work function of given material.
The initial kinetic of the photoelectron is \[K.E.=h\upsilon -\phi \].
\[\Rightarrow h\upsilon =K.E+\phi \] --- (1)
Now, kinetic energy after the increase in frequency will be,
\[K.E'=2h\upsilon -\phi \]
We will substitute the value of \[h\upsilon \] in this equation.
\[\begin{align}
 & \Rightarrow K.E'=2\left( K.E+\phi \right)-\phi \\
 & \Rightarrow K.E'=2K.E+2\phi -\phi \\
 & \Rightarrow K.E'=2K.E+\phi \\
\end{align}\]
From this, we found that kinetic energy increases by more than double.

So, option c is the correct answer.

Note:
We must be aware that in the case of photoelectric effect the ejected electrons are called photoelectrons, but their behavior and properties are exactly like normal electrons. This question can be twisted and asked as the change in wavelength also. So, as we know the relation between frequency and wavelength, we can easily answer that too. Frequency is inversely proportional to wavelength.