Question

The threshold wavelength of Lithium is 8000 ${A^\circ }$. When light of wavelength 9000 ${A^\circ }$ is made to be incident on it, what happens to the photo electrons?
(A) They will not be emitted
(B) They will be emitted
(C) They will sometimes be emitted and sometimes not
(D) Data insufficient

Answer 1

Hint
When light with high frequency is incident on a metal plate, it tends to excite the electrons in it. The emission of electrons, however, is fully dependent on the threshold frequency of the metal in question.
$\Rightarrow v\alpha \dfrac{1}{\lambda }$, where $v$ is the frequency of radiation and $\lambda $ is the wavelength of radiation.

Complete step by step answer
Photoelectric emission is the phenomenon through which the free electrons are liberated from the surface of metals when it absorbs light. The electrons relieved in this manner are called photoelectrons. The photoelectric emission is dependent on the frequency of light and not on the intensity of light. The emission is only possible when light or electromagnetic radiation of a frequency higher than the threshold frequency falls on the surface of the metal.
In this question, we are asked to find if the given light would be able to cause photoelectric emission. The data provided to us is:
Threshold wavelength ${\lambda _0} = 8000{A^\circ }$
Incident wavelength $\lambda = 9000{A^\circ }$
We know that the frequency of light is inversely proportional to its wavelength. This gives us:
$\Rightarrow v\alpha \dfrac{1}{\lambda }$
Since $\lambda > {\lambda _0}$, due to the inverse relation:
$\Rightarrow v < {v_0}$
This implies the incident light has a frequency lower than the threshold. Thus, no electrons will be emitted.
Hence, the answer is option (A).

Note
The effect of photoelectric emission was discovered by Albert Einstein. He also received a Nobel Prize for this discovery. This theory led the foundation of the quantum nature of light and electrons and influenced the formation of the concept of wave–particle duality.