A graph is plotted with the de Broglie wavelength \( \lambda \) of an electron along the y-axis and the reciprocal of its momentum, \( 1/p \), along the x-axis. If \( h \) is Planck’s constant, the slope of the graph is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( h \)
The de Broglie relation gives the wavelength of a moving particle in terms of its momentum alone, \( \lambda=h/p \). With \( 1/p \) taken as the variable, it reads
\( \displaystyle \lambda=h\times\frac{1}{p} \)
This has the form \( y=kx \), with \( y=\lambda \) and \( x=1/p \). So the graph of \( \lambda \) against \( 1/p \) is a straight line through the origin. Its slope is the constant that multiplies \( 1/p \), which is Planck’s constant \( h \) itself, with no factor of \( 2\pi \). The mass of the electron does not enter, because the wavelength depends only on the momentum.
Therefore, the slope of the graph is \( h \).
NCERT (2026-27):
Summary – “9. The de Broglie wavelength (λ) associated with a moving particle is related to its momentum p as: λ = h/p. The dualism of matter is inherent in the de Broglie relation which contains a wave concept (λ) and a particle concept (p). The de Broglie wavelength is independent of the charge and nature of the material particle.”