A particle has kinetic energy \( K \) and de Broglie wavelength \( \lambda \). If \( h \) is Planck’s constant, the speed of the particle is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{2K\lambda}{h} \)
By the de Broglie relation, a particle of wavelength \( \lambda \) has momentum \( p=h/\lambda \). Since \( p=mv \), the kinetic energy can also be written in terms of the momentum:
\( \begin{aligned} K &= \frac{1}{2}mv^{2} \\ &= \frac{1}{2}(mv)\,v \\ &= \frac{1}{2}pv \end{aligned} \)
The speed of the particle is then:
\( \begin{aligned} v &= \frac{2K}{p} \\ &= \frac{2K}{h/\lambda} \\ &= \frac{2K\lambda}{h} \end{aligned} \)
The mass of the particle is not needed, since \( K \) and \( \lambda \) together fix the speed.