The correct answer is (D) : 3.42 eV
Frequency of EM waves
\(= \frac{6}{2π} × 10^{15}\)
and
\(\frac{9}{2π} × 10^{15}\)
Energy of one photon of these waves
\(=(4.14 × 10^{-15} × \frac{6}{2π} × 10^{15})eV\)
and
\((4.14 × 10^{-15} × \frac{9}{2π} ×10^{15})eV\)
3.95 eV and 5.93 eV
⇒ Energy of maximum energetic electrons
= 5.93 – 2.50
= 3.43 eV
Which of the following statements are correct, if the threshold frequency of caesium is $ 5.16 \times 10^{14} \, \text{Hz} $?
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is:
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,