Answer
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Hint: In order to solve this question, we will first calculate the de Broglie wavelength for an electron of given mass and velocity and then de Broglie wavelength for a proton and then we will find the required ratio of their wavelengths.
Formula Used:
The de Broglie wavelength is calculated using the formula:
$\lambda = \dfrac{h}{{mv}}$
where
h-The Planck’s constant
m-The mass of a particle and
v - The velocity of the particle
Complete answer:
We have given that, an electron and proton are moving with the same speed, let us assume their velocity is $v$ and the relation between the mass of the electron and proton is given as ${m_p} = 1836{m_e}$
Now, the de Broglie wavelength for an electron is calculated using the formula
$\lambda = \dfrac{h}{{mv}}$ we get,
${\lambda _{electron}} = \dfrac{h}{{{m_e}v}} \to (i)$
The de Broglie wavelength for proton is calculated as ${\lambda _{proton}} = \dfrac{h}{{{m_p}v}} \to (ii)$
Now, divide the equation (i) by (ii) we get,
$
\dfrac{{{\lambda _{electron}}}}{{{\lambda _{proton}}}} = \dfrac{{\dfrac{h}{{{m_e}v}}}}{{\dfrac{h}{{{m_p}v}}}} \\
\dfrac{{{\lambda _{electron}}}}{{{\lambda _{proton}}}} = \dfrac{{{m_p}}}{{{m_e}}} \\
$
Since, we have given that ${m_p} = 1836{m_e}$ or by rearranging it we have $\dfrac{{{m_p}}}{{{m_e}}} = 1836$
Substituting the above value in the equation we get:
$\dfrac{{{\lambda _{electron}}}}{{{\lambda _{proton}}}} = \dfrac{{{m_p}}}{{{m_e}}}$
$\Rightarrow \dfrac{{{\lambda _{electron}}}}{{{\lambda _{proton}}}} = 1836$
Therefore, the ratio of the wavelength of electron and proton is 1836.
Hence, the correct option is (B) 1836
Note: It should be remembered that de Broglie wavelength is the wavelength associated with all the particles however larger bodies having larger mass show a very negligible amount of wave nature whereas for elementary particles this effect is very noticeable.
Formula Used:
The de Broglie wavelength is calculated using the formula:
$\lambda = \dfrac{h}{{mv}}$
where
h-The Planck’s constant
m-The mass of a particle and
v - The velocity of the particle
Complete answer:
We have given that, an electron and proton are moving with the same speed, let us assume their velocity is $v$ and the relation between the mass of the electron and proton is given as ${m_p} = 1836{m_e}$
Now, the de Broglie wavelength for an electron is calculated using the formula
$\lambda = \dfrac{h}{{mv}}$ we get,
${\lambda _{electron}} = \dfrac{h}{{{m_e}v}} \to (i)$
The de Broglie wavelength for proton is calculated as ${\lambda _{proton}} = \dfrac{h}{{{m_p}v}} \to (ii)$
Now, divide the equation (i) by (ii) we get,
$
\dfrac{{{\lambda _{electron}}}}{{{\lambda _{proton}}}} = \dfrac{{\dfrac{h}{{{m_e}v}}}}{{\dfrac{h}{{{m_p}v}}}} \\
\dfrac{{{\lambda _{electron}}}}{{{\lambda _{proton}}}} = \dfrac{{{m_p}}}{{{m_e}}} \\
$
Since, we have given that ${m_p} = 1836{m_e}$ or by rearranging it we have $\dfrac{{{m_p}}}{{{m_e}}} = 1836$
Substituting the above value in the equation we get:
$\dfrac{{{\lambda _{electron}}}}{{{\lambda _{proton}}}} = \dfrac{{{m_p}}}{{{m_e}}}$
$\Rightarrow \dfrac{{{\lambda _{electron}}}}{{{\lambda _{proton}}}} = 1836$
Therefore, the ratio of the wavelength of electron and proton is 1836.
Hence, the correct option is (B) 1836
Note: It should be remembered that de Broglie wavelength is the wavelength associated with all the particles however larger bodies having larger mass show a very negligible amount of wave nature whereas for elementary particles this effect is very noticeable.
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