Question

If the work function of the metal is W and the frequency is $\nu$ , then there is no emission of photoelectrons if:
$\begin{aligned} & (A)v\langle \dfrac{W}{h} \\\ & (B)v\rangle \dfrac{W}{h} \\\ & (C)v\ge \dfrac{W}{h} \\\ & (D)v\le \dfrac{W}{h} \\\ \end{aligned}$

Answer 1

In photoelectric effect, the photoelectrons are emitted when light waves of certain wavelengths hit the metal surface. This energy should at least be greater than the work function of the metal used. So, in order to just stop the emission of photoelectrons, the incident energy should be just enough that the photoelectrons do not acquire any kinetic energy of their own.

Complete answer:
Let a light beam of frequency $v$ be struck on the metal surface used for photoelectric effect.
And, let the energy corresponding to this light beam be E. Then, E will be equal to:
$\Rightarrow E=hv$
Where, ‘h’ is the Planck’s constant.
Now, ideally for the emission of photoelectrons, we can write:
$\Rightarrow E=W+K.E.$
Where, K.E. is the kinetic energy of the emitted photoelectrons.
Now, in order to just stop the photoelectric emission, the photoelectrons should not move from their respective orbits in their respective atoms. That is, its velocity should be zero. This implies that the kinetic energy of the photoelectrons should be zero.
$\Rightarrow K.E.=0$
Now, if the energy provided to the atoms in the metal is just less than the work function of the metal, then we would get the limiting condition for no photoelectron emission. Mathematically, this could be written as:
$\begin{aligned} & \Rightarrow E\langle W \\\ & \Rightarrow hv\langle W \\\ & \therefore v\langle \dfrac{W}{h} \\\ \end{aligned}$
Hence, the condition for which, if the work function of the metal is W and the frequency is $\nu$ there is no emission of photoelectrons is $v\langle \dfrac{W}{h}$.

Hence, option (A) is the correct option.

Note:
We haven’t taken into consideration the less-than equal-to sign because, at that condition the photoelectrons have got just energy to escape. So, it would be ambiguous to comment in such a scenario. That is why we took only the less than sign which is totally valid for all the cases and has no ambiguity.