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Step-by-Step Solution
Step 1: Understand the Problem
An electron is accelerated by a potential $V_1$, giving it a de Broglie wavelength $\lambda$. When the accelerating potential changes to $V_2$, the new de Broglie wavelength becomes $1.5\lambda$ (that is, it increases by 50%). We want to find the ratio $V_1 / V_2$.
Step 2: Recall the Relation Between Electron Momentum and Accelerating Potential
An electron accelerated through a potential difference $V$ gains kinetic energy equal to the electric potential energy $eV$. Its momentum $P$ can be written as:
$P = \sqrt{2\, e\, V\, m}$
where:
$e$ is the charge of the electron
$m$ is the mass of the electron
$V$ is the accelerating potential difference
Step 3: Express the de Broglie Wavelength
The de Broglie wavelength $\lambda$ is given by:
$\lambda = \frac{h}{P}$
where $h$ is the Planck's constant.
Step 4: Write the Wavelengths for the Two Potentials
For the first potential $V_1$:
$\lambda = \frac{h}{P_1} = \frac{h}{\sqrt{2\,e\,V_1\,m}}$
For the second potential $V_2$, the new wavelength is $1.5 \lambda$. Hence,
$1.5 \lambda = \frac{h}{P_2} = \frac{h}{\sqrt{2\,e\,V_2\,m}}$
Step 5: Form the Ratio of Momenta
From the two expressions, dividing the first by the second:
$\frac{\lambda}{1.5 \lambda} \;=\; \frac{\frac{h}{\sqrt{2\,e\,V_1\,m}}}{\frac{h}{\sqrt{2\,e\,V_2\,m}}}
\;\Longrightarrow\; \frac{1}{1.5} = \frac{\sqrt{2\,e\,V_2\,m}}{\sqrt{2\,e\,V_1\,m}}
\;\Longrightarrow\; \frac{2}{3} = \sqrt{\frac{V_2}{V_1}}.
$
Step 6: Solve for the Ratio of Potentials
Squaring both sides of
$\frac{2}{3} = \sqrt{\frac{V_2}{V_1}}$:
$\left(\frac{2}{3}\right)^2 = \frac{V_2}{V_1}
\;\Longrightarrow\; \frac{4}{9} = \frac{V_2}{V_1}
\;\Longrightarrow\; \frac{V_1}{V_2} = \frac{9}{4}.
$
Step 7: Conclusion
The required ratio $V_1 / V_2$ is
$\frac{9}{4}$,
which matches the given correct answer.
