KCET · Physics · Dual Nature of Matter
The additional energy that should be given to an electron to reduce its de-Broglie wavelength from \(1 \mathrm{~nm}\) to \(0.5 \mathrm{~nm}\) is
- A 2 times the initial kinetic energy
- B 3 times the initial kinetic energy
- C \(0.5\) times the initial kinetic energy
- D 4 times the initial kinetic energy
Answer & Solution
Correct Answer
(B) 3 times the initial kinetic energy
Step-by-step Solution
Detailed explanation
The de-Broglie wavelength,
\(\lambda=\frac{h}{\sqrt{2 m E_{k}}} \quad \therefore \lambda \propto \frac{1}{\sqrt{E_{k}}}\)
( \(\because\) h and \(m\) remains constant)
Here, in given condition
\(\begin{array}{ll}
\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{E_{k_{2}}}{E_{k_{1}}}}, \quad \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{4 E}{E}} \\
\lambda_{1}=2 \lambda_{2}, & \lambda_{2}=\frac{\lambda_{1}}{2}
\end{array}\)
So, if \(E_{k}\) is increased by four times, then \(\lambda\) becomes half.
\(\therefore\) Additional kinetic energy that should be supplied to the electron
\(=4 E_{k}-E_{k}=3 E_{k}\)
\(\lambda=\frac{h}{\sqrt{2 m E_{k}}} \quad \therefore \lambda \propto \frac{1}{\sqrt{E_{k}}}\)
( \(\because\) h and \(m\) remains constant)
Here, in given condition
\(\begin{array}{ll}
\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{E_{k_{2}}}{E_{k_{1}}}}, \quad \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{4 E}{E}} \\
\lambda_{1}=2 \lambda_{2}, & \lambda_{2}=\frac{\lambda_{1}}{2}
\end{array}\)
So, if \(E_{k}\) is increased by four times, then \(\lambda\) becomes half.
\(\therefore\) Additional kinetic energy that should be supplied to the electron
\(=4 E_{k}-E_{k}=3 E_{k}\)
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