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Step-by-Step Solution
Step 1: Identify the Relation Between Kinetic Energy and Accelerating Potential
When a charged particle of charge $q$ is accelerated by a potential difference $V$, it acquires a kinetic energy given by:
$K = q \, V$
Step 2: Write the de Broglie Wavelength Formula
The de Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ (accelerated from rest by a potential $V$) is given by:
$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m K}} = \frac{h}{\sqrt{2 m q V}}$
where $h$ is Planck’s constant, and $p$ is the linear momentum of the particle.
Step 3: Express the Ratio of Wavelengths
For particle A (mass $m_A = m$, charge $q_A = q$, and potential $V_A = 50\,\text{V}$) and particle B (mass $m_B = 4m$, charge $q_B = q$, and potential $V_B = 2500\,\text{V}$), their de Broglie wavelengths are:
$\lambda_A = \frac{h}{\sqrt{2\,m_A\,q_A\,V_A}}$
$\lambda_B = \frac{h}{\sqrt{2\,m_B\,q_B\,V_B}}$
The ratio of their wavelengths is:
$\frac{\lambda_A}{\lambda_B} = \frac{\sqrt{2\,m_B\,q_B\,V_B}}{\sqrt{2\,m_A\,q_A\,V_A}}$
Step 4: Substitute the Given Values
Substitute $m_B = 4m$, $q_B = q$, $V_B = 2500$, $m_A = m$, $q_A = q$, and $V_A = 50$:
$\frac{\lambda_A}{\lambda_B}
= \sqrt{\frac{(4m)\cdot q \cdot 2500}{m \cdot q \cdot 50}}
= \sqrt{\frac{4 \times 2500}{50}}
$
Step 5: Simplify the Expression
$\frac{4 \times 2500}{50} = \frac{10000}{50} = 200
$
$\sqrt{200} = \sqrt{100 \times 2} = 10 \times \sqrt{2} \approx 14.14
$
Step 6: State the Final Answer
The ratio of de Broglie wavelengths ${\lambda_A}/{\lambda_B}$ is approximately 14.14.
