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Question

$$Be^{3+}$$ and a proton are accelerated by the same potential, their de-Broglie wavelenghts have the ratio (assume mass of proton = mass of neutron):


A
1:2
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B
1:4
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C
1:1
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D
1:33
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Solution

The correct option is C $$1:3\sqrt{3}$$
The de-Broglie wavelength for a mass $$m$$ having a charge $$q$$ when accelerated by a potential difference $$V$$ is given by,
$$\lambda _b=\dfrac{h}{\sqrt{2mqV}}$$
For the same potential,
$$\lambda_b \propto \dfrac{1}{\sqrt{qm}}$$
For proton, let $$q_1=q_p=e$$ and $$m_1=m_p=m$$
Hence, $$q_2=q_{Be^{3+}}=3e$$ and $$m_2=m_{Be^{3+}}=9m$$

Therefore,
$$\dfrac{\lambda _2}{\lambda_1}=\dfrac{\sqrt{me}}{\sqrt{27me}}=\dfrac{1}{3\sqrt{3}}$$

Chemistry

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