Question 351: Dual Nature of Matter and Radiation

Question

According to de-Broglie hypothesis if an electron of mass '$$m$$' is accelerated by potential difference '$$V$$', the associated wavelength is '$$\lambda$$'. When a proton of mass '$$\mathrm{M}$$' is accelerated through potential difference $$9 \mathrm{~V}$$, then the wavelength associated with it is

Accepted Answer

$$\lambda=\frac{h}{\sqrt{2 \mathrm{mqV}}}$$
For electron and proton change $$\mathrm{q}$$ has same value. If $$\lambda$$' is the wavelength of the proton, then
$$\begin{aligned}
& \lambda^{\prime}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{Mq} 	imes 9 \mathrm{v}}} \
& 	herefore \frac{\lambda^{\prime}}{\lambda}=\frac{1}{3} \sqrt{\frac{\mathrm{m}}{\mathrm{M}}}
\end{aligned}$$

Answer

$$\frac{\lambda}{3} \sqrt{\frac{\mathrm{m}}{\mathrm{M}}}$$

Answer

$$\frac{3}{2} \sqrt{\frac{m}{M}}$$

Answer

$$\frac{\lambda}{3} \sqrt{\frac{M}{m}}$$

Answer

$$\frac{3}{\lambda} \sqrt{\frac{\mathrm{M}}{\mathrm{m}}}$$

Dual Nature of Matter and Radiation

According to de-Broglie hypothesis if an electron of mass '$m' is accelerated by potential difference 'V', the associated wavelength is '\lambda'. When a proton of mass '\mathrm{M}' is accelerated through potential difference 9 \mathrm{~V}$, then the wavelength associated with it is

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