Work function \(W =4 eV =4 \times 1.6 \times 10^{-19} J\) \(=6.4 \times 10^{-19} J\) Wavelength of incident radiation \(\lambda=0.2 \mu m =0.2 \times 10^{-6} m\) Maximum KE of liberated electron \(( KE )_{\max }=\frac{h c}{\lambda}-W\) \(\frac{1}{2} m v_{\max }^{2}=\frac{h c}{\lambda}-W\) \(\frac{1}{2} \times 9.1 \times 10^{-31} v_{\max }^{2}\) \(=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{0.2 \times 10^{-6}}-6.4 \times 10^{-19}\) \(=9.9 \times 10^{-19}-6.4 \times 10^{-19}=3.5 \times 10^{-19}\) \(\therefore v_{\max }=\sqrt{\frac{3.5 \times 10^{-19} \times 2}{9.1 \times 10^{-31}}}=\sqrt{\frac{7}{9.1} \times 10^{12}}\) \(=8.8 \times 10^{5} m / s\)
The Photoelectric Effect is a phenomena in which an electron is emitted from a metallic surface when a beam of light touches it. Photoelectrons are the electrons that are released from the surface.
Given below are two statements: one is labelled as Assertion (A) and the other one is labelled as Reason (R).
Assertion (A): Emission of electrons in the photoelectric effect can be suppressed by applying a sufficiently negative electron potential to the photoemissive substance.
Reason (R): A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with the frequency of incident radiation.
In light of the above statements, choose the most appropriate answer from the options given below:
When light shines on a metal, electrons can be ejected from the surface of the metal in a phenomenon known as the photoelectric effect. This process is also often referred to as photoemission, and the electrons that are ejected from the metal are called photoelectrons.
According to Einstein’s explanation of the photoelectric effect :
The energy of photon = energy needed to remove an electron + kinetic energy of the emitted electron
i.e. hν = W + E
Where,