Question

# A proton and an alpha particle are accelerated through the same potential difference. The ratio of wavelengths associated with proton and alpha particle respectively is :

A
1:22
B
2:1
C
22:1
D
4:1

Solution

## The correct option is C $$2\sqrt{2}:1$$We know, $$\dfrac{1}{2}mv^2=eV$$$$mv = \sqrt{2meV}$$$$wavelength, \ \lambda = \dfrac{h}{mv}$$$$\lambda _p = \dfrac{h}{\sqrt{2m_{p}q_{p}V}}$$$$\lambda _p = \dfrac{h}{\sqrt{2meV}}$$$$\lambda _{\alpha} = \dfrac{h}{\sqrt{2m_{\alpha} q_{\alpha}V}}$$$$= \dfrac{h}{\sqrt{2\times4m\times2e\times V}}$$$$\dfrac{\lambda _p}{\lambda _{\alpha }} = \dfrac{2\sqrt2}{1}$$So, the answer is option (C).Physics

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