TS EAMCET · Physics · Dual Nature of Matter
The de Broglie wavelength of a particle moving with a speed of \(0.8 \mathrm{c}\) is equal to the wavelength of a photon. If \(\mathrm{c}\) is speed of the photon in vacuum, the ratio of the energy of the photon and the kinetic energy of the particle is
- A \(2: 3\)
- B 5:2
- C 4:5
- D 3:5
Answer & Solution
Correct Answer
(B) 5:2
Step-by-step Solution
Detailed explanation
(b) de Brogle wavelength of a particle \(=\) wavelength of a photon \[ \begin{aligned} & \frac{\mathrm{h}}{\mathrm{mv}}=\lambda \\ & \Rightarrow \lambda=\frac{\mathrm{h}}{0.8 \mathrm{~cm}} \end{aligned} \] Energy of photon \(=\frac{\mathrm{hc}}{\lambda}\) Kinetic energy of the…
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-
Momentum before collision,
\(
p_1=4 m v_1-2 m v_2
\)
Momentum after collision,
\(
p_2=6 m\left(-5 v_2\right)=-30 m v_2
\)
Conservation of momentum gives,
\(
\begin{gathered}
p_1=p_2 \\
\Rightarrow \quad 4 m v_1-2 m v_2=-30 m v_2 \quad \Rightarrow 4 v_1=-32 v_2 \\
\Rightarrow \quad \frac{v_1}{v_2}=\frac{-32}{4}=-8 \text { or }\left|\frac{v_1}{v_2}\right|=8
\end{gathered}
\)TS EAMCET 2018 Hard