The de Broglie wavelength of a proton of kinetic energy \( K \) is \( \lambda_0 \). Match List-I with List-II, where List-II gives the de Broglie wavelength of each particle. (Take the masses of the deuteron and the alpha particle as 2 and 4 times the mass of the proton.)
| List-I | List-II |
| (A) A proton of kinetic energy \( 4K \) | (I) \( \lambda_0/4 \) |
| (B) A deuteron of kinetic energy \( K \) | (II) \( 2\lambda_0 \) |
| (C) An alpha particle of kinetic energy \( 4K \) | (III) \( \lambda_0/\sqrt{2} \) |
| (D) A proton of kinetic energy \( K/4 \) | (IV) \( \lambda_0/2 \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
For a particle of mass \( m \) and kinetic energy \( K \), the momentum is \( p=\sqrt{2mK} \). So the de Broglie wavelength is
\( \displaystyle \lambda=\frac{h}{\sqrt{2mK}} \)
The wavelength then varies as \( 1/\sqrt{mK} \). Each case is compared with the proton of kinetic energy \( K \), for which \( \lambda=\lambda_0 \).
(A) A proton of kinetic energy \( 4K \) — This matches (IV) \( \lambda_0/2 \). The mass is unchanged and the energy is 4 times larger, so \( mK \) is 4 times larger and the wavelength is halved.
(B) A deuteron of kinetic energy \( K \) — This matches (III) \( \lambda_0/\sqrt{2} \). The deuteron has twice the mass of the proton and the same energy, so \( mK \) is doubled. The wavelength becomes \( \lambda_0/\sqrt{2} \).
(C) An alpha particle of kinetic energy \( 4K \) — This matches (I) \( \lambda_0/4 \). Both the mass and the energy are 4 times larger, so \( mK \) grows 16 times. The wavelength becomes \( \lambda_0/\sqrt{16} \), that is \( \lambda_0/4 \).
(D) A proton of kinetic energy \( K/4 \) — This matches (II) \( 2\lambda_0 \). With a quarter of the energy, \( mK \) drops to a quarter, so the wavelength doubles.