An electron beam and a proton beam have the same energy per particle. The wavelength associated with the electron beam is 24 A $^{\circ}$ .The wavelength associated with the proton beam would approximately be

200 A

^{\circ}

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10 A

^{\circ}

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24 A

^{\circ}

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0.5 A

^{\circ}

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Solution

The correct option is B 0.5 A $^{\circ}$
De Broglie's wavelength $λ = \frac{h}{m v} = \frac{h}{\sqrt{2 m E}}$

For same energy $E$ per particle $\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{m_{e}}{m_{p}}} = \sqrt{\frac{1}{1840}}$

$λ_{p} = λ_{e} \frac{1}{42.89} = \frac{24}{42.9} = 0.559 ≅ 0.5 A^{o}$ .

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An electron beam and a proton beam have the same energy per particle. The wavelength associated with the electron beam is 24 A∘.The wavelength associated with the proton beam would approximately be

The correct option is B 0.5 A∘De Broglie's wavelength λ=hmv=h√2mEFor same energy E per particle λpλe=√memp=√11840 λp=λe142.89=2442.9=0.559≅0.5Ao.

An electron beam and a proton beam have the same energy per particle. The wavelength associated with the electron beam is 24 A $^{\circ}$ .The wavelength associated with the proton beam would approximately be

The correct option is B 0.5 A $^{\circ}$
De Broglie's wavelength $λ = \frac{h}{m v} = \frac{h}{\sqrt{2 m E}}$

For same energy $E$ per particle $\frac{λ_{p}}{λ_{e}} = \sqrt{\frac{m_{e}}{m_{p}}} = \sqrt{\frac{1}{1840}}$

$λ_{p} = λ_{e} \frac{1}{42.89} = \frac{24}{42.9} = 0.559 ≅ 0.5 A^{o}$ .