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, $\mathrm{D}$ carrying same charge but of masses $\mathrm{m}_1, \mathrm{~m}_2$ , $\mathrm{m}_3, \mathrm{~m}_4$ . Which on represents a particle of largest mass?

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,

\mathrm{D}

carrying same charge but of masses

\mathrm{m}_1, \mathrm{~m}_2

,

\mathrm{m}_3, \mathrm{~m}_4

. Which on represents a particle of largest mass? Graph shows the variation of de-Broglie wavelength (λ) versus 1 V where ' V ' is the accelerating potential for four

Graph shows the variation of de-Broglie wavelength (λ) versus 1 V where ' V ' is the accelerating potential for four

Quantrex Academy · Accepted Answer

Correct answer: A. m_1

de Broglie wayelength λ= h p λ= h 2 mqv λ v = h 2 mq λ ( 1 v ) = 1 2 mq Slope of the graph = 1 2 mq The slope will be maximum for minimum mass. m_4 is minimum and m_1 will be minimum.

Graph shows the variation of de-Broglie wavelength (λ) versus 1 V...

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