KCET · Physics · Dual Nature of Matter
The de-Broglie wavelength of a particle of kinetic energy \(K\) is \(\lambda\), the wavelength of the particle, if its kinetic energy \(\frac{K}{4}\) is
- A \(2 \lambda\)
- B \(\frac{\lambda}{2}\)
- C \(4 \lambda\)
- D \(\lambda\)
Answer & Solution
Correct Answer
(A) \(2 \lambda\)
Step-by-step Solution
Detailed explanation
de-Broglie wavelength \((\lambda)\) of the particle in terms of kinetic energy \(K\) is given as
\[
\lambda=\frac{h}{\sqrt{2 m K}}
\]
where, \(m\) is the mass of the particle and \(h\) is Planck's constant.
Let at wavelength \(\lambda^{\prime}\), kinetic energy of particle becomes \(K^{\prime}=\frac{K}{4}\)
[from Eq. (i)]
\[
\lambda=\frac{h}{\sqrt{2 m K}}
\]
where, \(m\) is the mass of the particle and \(h\) is Planck's constant.
Let at wavelength \(\lambda^{\prime}\), kinetic energy of particle becomes \(K^{\prime}=\frac{K}{4}\)
[from Eq. (i)]
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