The deBroglie wavelength associated with a particle of mass $m$ and energy $E$ is $\frac{h}{\sqrt{2 m E}}$ . The dimensional formula of Planck's constant h is

Question

The deBroglie wavelength associated with a particle of mass m- moving with a velocity -v - and energy E is given bydfrac-h-mv-2-dfrac-mv-h-2-dfrac-h-sqrt-2mE-sqrt-2mE-h

Toppr Admin · Accepted Answer

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The deBroglie wavelength associated with a particle of mass m, moving with a velocity $^{'} v^{'}$ and energy E is given by
$\frac{h}{m v^{2}}$
$\frac{m v}{h^{2}}$
$\frac{h}{\sqrt{2 m E}}$
$\sqrt{2 m E} / h$

We know,
$\frac{1}{2} m v^{2} = E$
$m v = \sqrt{2 E m}$
so debroglie wavelength $(λ) = \frac{h}{m v}$
$= \frac{h}{\sqrt{2 E m}}$

The deBroglie wavelength associated with a particle of mass m, moving with a velocity ′v ′ and energy E is given byhmv2mvh2h√2mE√2mE/h

We know,12mv2=Emv=√2Emso debroglie wavelength (λ)=hmv =h√2Em

The deBroglie wavelength associated with a particle of mass m, moving with a velocity $^{'} v^{'}$ and energy E is given by
$\frac{h}{m v^{2}}$
$\frac{m v}{h^{2}}$
$\frac{h}{\sqrt{2 m E}}$
$\sqrt{2 m E} / h$

We know,
$\frac{1}{2} m v^{2} = E$
$m v = \sqrt{2 E m}$
so debroglie wavelength $(λ) = \frac{h}{m v}$
$= \frac{h}{\sqrt{2 E m}}$