Question

Light of wavelength \[3500{{A}^{0}}\] is incident on two metals 1 and 2 whose work function are $4.5eV\ \,And \,2.5eV$respectively then:
A. Both 1 and 2 emit photoelectrons
B. Only 2 emit photoelectrons
C. Only 1 emit photoelectrons
D. Neither 1 nor 2 emit photoelectrons

Answer 1

Hint:We are given the wavelength of the light in angstrom and this light is incident on two metals and we are given the work function of both the metals. In order to find out whether the given radiation of light will emit the photoelectrons or not we can make use of Einstein equation of photoelectric effect. The frequency corresponds to the frequency of incident radiation.

Complete step by step answer:
The wavelength is $3500{{A}^{0}}$, converting it into nanometers we get, $=\dfrac{3500}{10}=350nm$
Also, we know the energy of the incoming radiation can be found out as $E=\dfrac{hc}{\lambda }$
When we have the wavelength in nanometer then we can make use of the value of $hc=1240eVnm$
$\Rightarrow E=\dfrac{hc}{\lambda } \\
\Rightarrow E=\dfrac{1240}{350} \\
\therefore E=3.54eV $
The wok function of metal 1 is higher but the work function of metal 2 is lower. So, the electrons will get emitted from metal 2 and not from metal 1.

So, the correct option is B.

Note: Photo electric effect is the phenomenon of emission of electrons when radiation of a given frequency is incident on the metal. There exists a frequency called threshold frequency below which no photo electric effect can take place. Threshold frequency changes with the change in the material. Photoelectric equation can be written as \[h\nu =h{{\nu }_{0}}+e{{V}_{0}}\] where $hv$ is the energy of the incident radiation and \[h{{\nu }_{0}}\] is the work function of the metal.