Match List-I with List-II
| List-I | List-II |
| Quantity | Formula |
| (A) Einstein's photoelectric Equation | (I) \( K_{\max}/e \) |
| (B) de-Broglie wavelength | (II) \( K_{\max} = hc/\lambda - \Phi \) |
| (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)
Each item is a standard result of the photoelectric effect or of matter waves.
(A) Einstein's photoelectric Equation — this matches (II) \( K_{\max} = hc/\lambda - \Phi \). The photon energy \( hc/\lambda \) goes partly into the work function \( \Phi \) and the rest becomes the maximum kinetic energy.
(B) de-Broglie wavelength — this matches (IV) \( h/p \). A particle of momentum \( p \) has the wavelength \( \lambda = h/p \).
(C) Stopping potential — this matches (I) \( K_{\max}/e \). The retarding potential that just stops the fastest electrons satisfies \( eV_0 = K_{\max} \), so \( V_0 = K_{\max}/e \).
(D) Threshold wavelength — this matches (III) \( hc/\Phi \). At the threshold \( K_{\max} = 0 \), so \( hc/\lambda_0 = \Phi \) and \( \lambda_0 = hc/\Phi \).
NCERT (2026-27):
Summary – “… 4. … The stopping potential is directly related to the maximum kinetic energy of electrons emitted: \( eV_0 = (1/2)\,m v_{max}^2 = K_{max} \).”
Summary – “… 7. Einstein’s photoelectric equation is in accordance with the energy conservation law as applied to the photon absorption by an electron in the metal. The maximum kinetic energy \( (1/2)m v_{max}^2 \) is equal to the photon energy \( (h\nu) \) minus the work function \( \phi_0 \;(= h\nu_0) \) of the target metal:
\( \displaystyle \frac{1}{2} m v_{max}^2 = eV_0 = h\nu - \phi_0 = h(\nu - \nu_0) \)”
Summary – “… 9. The de Broglie wavelength \( (\lambda) \) associated with a moving particle is related to its momentum \( p \) as: \( \lambda = h/p \).”