Two positively charged particles m _ 1 and m _ 2 have been accelerated across the same potential difference of 200 keV as shown below. [Given mass of m _ 1 =1 amu and m _ 2 =4 amu] The deBroglie wavelength of m _ 1 will be x times of m _ 2 . The value of x is \_\_\_\_ (nearest integer)

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Correct answer: A. A

For a charged particle accelerated through potential V, the de Broglie wavelength is = h 2mVe . The wavelength is inversely proportional to m . Therefore: _1 _2 = m_2 m_1 = 4 1 = 2. This means _1 = 2 _2, so x = 2 (nearest integer).

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