
If the momentum of an electron is changed by $\Delta p$, then the de-Broglie wavelength associated will change by 0.50%. The initial momentum of electron will be
A) $\Delta p/200$
B) $\Delta p/100$
C) $200\Delta p$
D) $400\Delta p$
Answer
566.1k+ views
Hint: The question talks about generating momentum which indirectly involves the mass of the electron and the velocity or speed of the electron as we can now look out for the relationship that occurs between momentum and wavelength.
Complete step by step answer:
We are given to find the initial momentum of an electron when the momentum of the electron is changed by $\Delta p$ and the de-Broglie wavelength associated will change by 0.50%.
From de Broglie equation, where wavelength, Planck’s constant and momentum are connected together.
$\lambda = \dfrac{h}{p} \to \left( 1 \right)$
Where lambda is the wavelength ($\lambda $), h is the Planck’s constant and p is the momentum value.
Let the initial momentum be ${p_1}$ and the associated wavelength is ${\lambda _1}$
Also from the relation, the wavelength is indirectly proportional to the momentum, which means that as the wavelength increases, the momentum would decrease and as the wavelength decreases the momentum will also increase.
Hence $\Delta \lambda = \dfrac{{0.5}}{{100}} \times {\lambda _1}$ and $\Delta p = - P$
We would then have to take the derivative of the equation with respect to momentum, the equation then becomes
$\dfrac{{\Delta \lambda }}{{\Delta p}} = \dfrac{{{\lambda _1}}}{{{p_1}}}$
By substituting $\Delta \lambda = \dfrac{{0.5}}{{100}} \times {\lambda _1}$ into the above equation, the equation then we would arrive at will be $\dfrac{{0.5}}{{100}} \times \dfrac{{{\lambda _1}}}{{\Delta p}} = \dfrac{{{\lambda _1}}}{{{p_1}}}$
We would then make ${p_1}$ the subject of the relation where ${\lambda _1}$ will cancel out each other
$
{p_1} = \dfrac{{100}}{{0.5}} \times \Delta p \\
{p_1} = 200\Delta p \\
$
The answer is Option C, the initial momentum of electron will be $200\Delta p$
Note:The same happens, when we might be asked to find other parameters like wavelength, where there might be an increase or decrease, or momentum increase or decrease as well. This change occurs to both parameters.
Complete step by step answer:
We are given to find the initial momentum of an electron when the momentum of the electron is changed by $\Delta p$ and the de-Broglie wavelength associated will change by 0.50%.
From de Broglie equation, where wavelength, Planck’s constant and momentum are connected together.
$\lambda = \dfrac{h}{p} \to \left( 1 \right)$
Where lambda is the wavelength ($\lambda $), h is the Planck’s constant and p is the momentum value.
Let the initial momentum be ${p_1}$ and the associated wavelength is ${\lambda _1}$
Also from the relation, the wavelength is indirectly proportional to the momentum, which means that as the wavelength increases, the momentum would decrease and as the wavelength decreases the momentum will also increase.
Hence $\Delta \lambda = \dfrac{{0.5}}{{100}} \times {\lambda _1}$ and $\Delta p = - P$
We would then have to take the derivative of the equation with respect to momentum, the equation then becomes
$\dfrac{{\Delta \lambda }}{{\Delta p}} = \dfrac{{{\lambda _1}}}{{{p_1}}}$
By substituting $\Delta \lambda = \dfrac{{0.5}}{{100}} \times {\lambda _1}$ into the above equation, the equation then we would arrive at will be $\dfrac{{0.5}}{{100}} \times \dfrac{{{\lambda _1}}}{{\Delta p}} = \dfrac{{{\lambda _1}}}{{{p_1}}}$
We would then make ${p_1}$ the subject of the relation where ${\lambda _1}$ will cancel out each other
$
{p_1} = \dfrac{{100}}{{0.5}} \times \Delta p \\
{p_1} = 200\Delta p \\
$
The answer is Option C, the initial momentum of electron will be $200\Delta p$
Note:The same happens, when we might be asked to find other parameters like wavelength, where there might be an increase or decrease, or momentum increase or decrease as well. This change occurs to both parameters.
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