The ratio of de-Broglie wavelength of an $$\alpha$$-particle and a proton accelerated from rest by the same potential is $$\frac{1}{\sqrt{m}}$$, the value of $$m$$ is:

Solution :

De-Broglie wavelength is given by :

$$\lambda = \frac{h}{\sqrt{2mqV}}$$

For α-particle :

Mass,

$$m_\alpha = 4m_p$$

Charge,

$$q_\alpha = 2e$$

Therefore,

$$\lambda_\alpha = \frac{h}{\sqrt{2(4m_p)(2e)V}}$$

$$= \frac{h}{\sqrt{16m_peV}}$$

$$= \frac{h}{4\sqrt{m_peV}}$$

For proton :

$$\lambda_p = \frac{h}{\sqrt{2m_peV}}$$

Comparing with :

$$\frac{1}{\sqrt{m}}$$

we get,

$$m = 8$$

Final Answer :

$$8$$

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