Question

de-Broglie wavelength associated with an electron accelerated through a potential difference V is Lambda. what will be its wavelength when the accelerating potential is increased to 4v?

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Answer 1

$\bold {\huge {Answer :-}}$

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As we know from de-Broglie wavelength

$\Large\bold{\boxed{\boxed{\lambda\propto\:\frac{1}{\sqrt{V}}}}}$

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$\Large\bold{\therefore\:\frac{\lambda_1}{\lambda_2} =\sqrt{\frac{V_2}{V_1}}}$

Given $\Large\bold{V_2= 4V_1}$

$\Large\bold{\Rightarrow\:\frac{\lambda_1}{\lambda_2} =\sqrt{\frac{4V}{V}}}$

$\Large\bold{\Rightarrow\:\frac{\lambda_1}{\lambda_2} =2}$

$\Large\bold{\Rightarrow\:\lambda_2=\frac{\lambda_1}{2}}$

➛Hence, New wavelength is half of initial wavelength

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✭de-Broglie’s states, a moving material particle sometimes acts as a wave and sometimes as a particle; or a wave is associated with moving material particle, which controls the particle in every respect. The wave associated with moving particle is called matter wave or de Broglie wave, λ = h/(mv)