A proton is moving with speed \( v \). Consider the following particles:
(A) A deuteron moving with speed \( v/2 \)
(B) An alpha particle moving with speed \( v/2 \)
(C) A proton moving with speed \( v \) in the opposite direction
(D) A proton moving with speed \( 2v \)
How many of the above particles have the same de Broglie wavelength as the given proton? (Mass of the deuteron \( =2m_p \), mass of the alpha particle \( =4m_p \), where \( m_p \) is the mass of the proton)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Two
Two particles have the same de Broglie wavelength \( \lambda=h/p \) only when they have the same momentum, whatever their masses or charges. The given proton has momentum \( m_pv \), so each particle is checked against this value.
These particles have the same wavelength as the proton:
(A) Deuteron at \( v/2 \): its momentum is \( 2m_p\times v/2 \), that is \( m_pv \).
(C) Proton at \( v \) in the opposite direction: the size of its momentum is still \( m_pv \). The wavelength depends only on the size of the momentum. The direction of motion does not change it.
These do not:
(B) Alpha particle at \( v/2 \): its momentum is \( 4m_p\times v/2 \), that is \( 2m_pv \), so its wavelength is half that of the proton.
(D) Proton at \( 2v \): doubling the speed doubles the momentum to \( 2m_pv \), so the wavelength is halved.
Hence, two of the four particles have the same de Broglie wavelength as the given proton.
NCERT (2026-27):
Wave Nature of Matter – “… De Broglie proposed that the wave length λ associated with a particle of momentum p is given as
\( \displaystyle \lambda=\frac{h}{p} \)\( \displaystyle =\frac{h}{mv} \qquad (11.5) \)
where m is the mass of the particle and v its speed.”