Match List-I with List-II
|
List-I (Quantity) |
List-II (Formula) |
|
(A) Einstein's photoelectric Equation |
(I) $K_{max}/e$ |
|
(B) de-Broglie wavelength |
(II) $K_{max} =hc/λ-\phi$ |
|
(C) Stopping potential |
(III) $hc/\phi$ |
|
(D) Threshold wavelength |
(IV) $h/p$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
(A) Einstein's photoelectric Equation → (II) $K_{max} = \frac{hc}{\lambda} - \phi$
This is the fundamental relation for photoelectric emission, showing how the energy of an incident photon is used to overcome the work function ($\phi$) and the rest appears as kinetic energy ($K_{max}$) of the emitted electron.
(B) de-Broglie wavelength → (IV) $\frac{h}{p}$
According to de Broglie, any moving particle has a wavelength associated with it, inversely proportional to its momentum ($p$). This shows the wave nature of matter.
(C) Stopping potential → (I) $\frac{K_{max}}{e}$
The stopping potential is the potential needed to stop the fastest photoelectrons. It is related to their maximum kinetic energy by this formula.
(D) Threshold wavelength → (III) $\frac{hc}{\phi}$
The threshold wavelength is the longest wavelength (least energy) of light that can just cause photoemission. It's derived by setting $K_{max} = 0$ in Einstein’s equation.