Question 330: Dual Nature of Matter and Radiation

Question

Graph shows the variation of de-Broglie wavelength $$(\lambda)$$ versus $$\frac{1}{\sqrt{V}}$$ where '$$V$$' is the accelerating potential for four particles A, B, C, D carrying same charge but of masses $$\mathrm{m_1, m_2, m_3, m_4}$$. Which on represents a particle of largest mass?

Accepted Answer

de Broglie wavelength $$\lambda=\frac{h}{p}$$
$$\begin{aligned}
& 	herefore \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqv}}} \
& 	herefore \lambda \sqrt{\mathrm{v}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mq}}} \Rightarrow \frac{\lambda}{\left(\frac{1}{\sqrt{\mathrm{v}}}ight)}=\frac{1}{\sqrt{2 \mathrm{mq}}}
\end{aligned}$$
$$	herefore \quad$$ Slope of the graph $$=\frac{1}{\sqrt{2 \mathrm{mq}}}$$
The slope will be maximum for minimum mass,
$$	herefore \quad \mathrm{m}_4$$ is minimum and $$\mathrm{m}_1$$ will be minimum.

Answer

m$$_1$$

Answer

m$$_2$$

Answer

m$$_3$$

Answer

m$$_4$$

Dual Nature of Matter and Radiation

Graph shows the variation of de-Broglie wavelength $(\lambda) versus \frac{1}{\sqrt{V}} where 'V' is the accelerating potential for four particles A, B, C, D carrying same charge but of masses \mathrm{m_1, m_2, m_3, m_4}$. Which on represents a particle of largest mass?

Choose the correct answer: