Question

Light of wavelength $\lambda$ falls on a metal having work function $h_c$ / ${\lambda }_0$​. The photoelectric effect will take place only if

• A. $\lambda \le {\lambda }_0$
• B. $\lambda \ge 2{\lambda }_0$
• C. $\lambda <\raisebox{1ex}{${\lambda }_0$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• D. $\lambda \ge {\lambda }_0$

Answer 1

The correct option is A

$\lambda \le {\lambda }_0$

Wavelength:

It is the distance between two successive crests or two successive troughs of a wave, especially points in a sound wave or electromagnetic wave.

Work function:

It may be defined as that amount of minimum energy by which an electron can be ejected from any metal surface.

Explanation:

1. The electrons are ejected from the surface of a metal when the light of some wavelength strikes the surface of the metal, this effect is called the Photoelectric effect.
2. It will take place only if the energy of striking light will be greater than or equal to the work function of the metal.
3. We know $E=\frac{hc}{\lambda }$, Where $E$=energy of the photon. $h$=plank's constant, $c$= velocity of light. $\lambda$=wavelength of the photon,
4. As per the condition of the photoelectric effect, it can be written that

$$\begin{gathered} \Rightarrow \frac{hc}{\lambda }\ge \frac{hc}{{\lambda }_0} \\ or\frac{1}{\lambda }\ge \frac{1}{{\lambda }_0} \\ \Rightarrow \lambda \le {\lambda }_0 \end{gathered}$$

Hence, the Photoelectric effect will take place only if $\lambda \le {\lambda }_0$

Thus, the correct option is option A.