Assertion (A) : If an electron has the same wavelength as a photon, they have the same energy.
Reason (R) : By the de Broglie hypothesis, \( p=h/\lambda \) for both the electron and the photon.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (R) is true but (A) is false.
Assertion (A) If an electron has the same wavelength as a photon, they have the same energy. — False. The same wavelength gives the same momentum \( p=h/\lambda \), but energy depends on momentum differently for the photon and for the electron (mass \( m \), speed \( v \ll c \)):
\( E_{\mathrm{photon}}=pc \)
\( \displaystyle E_{\mathrm{electron}}=\frac{p^{2}}{2m} \)
\( \begin{aligned} \frac{E_{\mathrm{electron}}}{E_{\mathrm{photon}}} &= \frac{p}{2mc} \\ &= \frac{v}{2c} \end{aligned} \)
Since \( v \) is far less than \( c \), the electron has much less energy than the photon of the same wavelength.
Reason (R) By the de Broglie hypothesis, \( p=h/\lambda \) for both the electron and the photon. — True. The de Broglie relation \( \lambda=h/p \) gives the wavelength of a material particle such as the electron. A photon has \( p=h\nu/c \), so it satisfies the same relation, \( p=h/\lambda \).
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”
\( \displaystyle \lambda=\frac{h}{p} \)\( \displaystyle =\frac{h}{mv} \qquad (11.5) \)
Wave Nature of Matter – “Equation (11.5) for a material particle is basically a hypothesis whose validity can be tested only by experiment. However, it is interesting to see that it is satisfied also by a photon. For a photon, as we have seen,”
\( p=h\nu/c \qquad (11.6) \)
“Therefore,”
\( \displaystyle p=\frac{h\nu}{c}=\frac{h}{\lambda} \qquad (11.7) \)