Two positively charged particles \(m_1\) and \(m_2\) have been accelerated across the same potential difference of 200 keV. Given mass of \(m_1 = 1 \,\text{amu}\) and \(m_2 = 4 \,\text{amu}\). The de Broglie wavelength of \(m_1\) will be \(x\) times that of \(m_2\). The value of \(x\) is _______ (nearest integer).
Two positively charged particles \(m_1\) and \(m_2\) have been accelerated across the same potential difference of 200 keV. Given mass of \(m_1 = 1 \,\text{amu}\) and \(m_2 = 4 \,\text{amu}\). The de Broglie wavelength of \(m_1\) will be \(x\) times that of \(m_2\). The value of \(x\) is _______ (nearest integer).
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Correct Answer: 2
Solution and Explanation
Step 1: Relation of wavelengths. Since both particles are accelerated through the same potential difference, they gain the same kinetic energy \(E\). \[ \lambda \propto \frac{1}{\sqrt{m}} \]
Step 2: Ratio of wavelengths. \[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2}{m_1}} \]
Step 3: Substitute values. \[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{4}{1}} = 2 \] Final Answer: \[ x = 2 \]
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